In the process of extraction of gold, Roasted gold ore $ + {\rm{C}}{{\rm{N}}^ - }{\rm{\;}} + {\rm{\;}}{{\rm{H}}_2}{\rm{O\;}}\mathop \to \limits^{{{\rm{O}}_2}} {\rm{\;}}\left[ X \right]{\rm{\;}} + {\rm{\;O}}{{\rm{H}}^ - }$ $\left[ X \right]{\rm{\;}} + {\rm{\;Zn\;}} \to {\rm{\;}}\left[ Y \right]{\rm{\;}} + {\rm{\;Au}}$ Identify the complexes [X] and [Y]

$X = {\left[ {{\rm{Au}}{{\left( {{\rm{CN}}} \right)}_2}} \right]^ - },{\rm{\;}}Y = {\left[ {{\rm{Zn}}{{\left( {{\rm{CN}}} \right)}_4}} \right]^{2 - }}$

$X = {\left[ {{\rm{Au}}{{\left( {{\rm{CN}}} \right)}_4}} \right]^{3 - }},{\rm{\;}}Y = {\left[ {{\rm{Zn}}{{\left( {{\rm{CN}}} \right)}_4}} \right]^{2 - }}$

$X = {\left[ {{\rm{Au}}{{\left( {{\rm{CN}}} \right)}_2}} \right]^ - },{\rm{\;}}Y = {\left[ {{\rm{Zn}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{4 - }}$

$X = {\left[ {{\rm{Au}}{{\left( {{\rm{CN}}} \right)}_4}} \right]^ - },{\rm{\;}}Y = {\left[ {{\rm{Zn}}{{\left( {{\rm{CN}}} \right)}_4}} \right]^{2 - }}$

${\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3}$

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS MCQs for NEET UG CHEMISTRY — Hard

1 Hard

1 of 32Question permalink

Remembering +4 | -1

Mac Arthur and Forest cyanide process is used in the extraction of:

View answer and explanation

Correct option: B

Ag and Au

Option B (Ag and Au) is correct because the Mac Arthur and Forest cyanide process is specifically designed for the extraction of silver and gold from their ores.

Solution approach

Correct Answer + Core Reason: Ag and Au. The cyanide process is used for leaching silver and gold ores.

Step-by-Step Solution:

Step 1: The Mac Arthur and Forest cyanide process involves leaching the ore with a dilute solution of sodium cyanide or potassium cyanide.

Step 2: This process selectively dissolves silver and gold, forming soluble cyanide complexes:
${\mathrm{4Au}} + {8\mathrm{CN}^ - } + {\mathrm{O}_2} + {2{\mathrm{H}_2}\mathrm{O}} \to 4{[{\mathrm{Au}{{\left( {\mathrm{CN}} \right)}_2}}}]^ - } + 4\mathrm{OH}^ -$
${\mathrm{4Ag}} + {8\mathrm{CN}^ - } + {\mathrm{O}_2} + {2{\mathrm{H}_2}\mathrm{O}} \to 4{[{\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}}}]^ - } + 4\mathrm{OH}^ -$

Step 3: The gold and silver are then recovered from the solution by reduction, typically using zinc dust.

Why other options are wrong:
- Option A: Cu (Copper) is extracted using different methods, such as froth flotation and electrolytic refining.
- Option C: Fe (Iron) is extracted using reduction in a blast furnace.
- Option D: Cr (Chromium) is extracted from chromite ore through a series of chemical processes, not the cyanide process.

Key Concepts:
- Extraction of metals
- Leaching
- Complex formation

Memory Aids:
- Cyanide process is specifically for gold and silver.

Quick Tips:
- Remember the specific extraction methods for common metals.

2 Hard

2 of 32Question permalink

Analyzing +4 | -1

Assertion (A): Wolframite impurities are separated from cassiterite by electromagnetic separation.

Reason (R): Cassiterite being magnetic is attracted by the magnet and forms a separate heap.

View answer and explanation

Correct option: C

(A) is true but (R) is false

Option C is correct because while the assertion is true, the reason is false. Wolframite is magnetic and is separated from non-magnetic cassiterite, not the other way around.

Solution approach

Correct Answer + Core Reason: (A) is true but (R) is false. Wolframite is magnetic and is separated from non-magnetic cassiterite.

Step-by-Step Solution:

Step 1: Analyze the assertion: Wolframite impurities are separated from cassiterite by electromagnetic separation.
- This is true. Electromagnetic separation is used to separate magnetic and non-magnetic substances.

Step 2: Analyze the reason: Cassiterite being magnetic is attracted by the magnet and forms a separate heap.
- This is false. Wolframite is magnetic, while cassiterite is non-magnetic.

Step 3: Determine the correctness of both statements.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Electromagnetic separation
- Magnetic properties of minerals

Memory Aids:
- Wolframite is magnetic.
- Cassiterite is non-magnetic.

Quick Tips:
- Remember the magnetic properties of common minerals.

3 Hard

3 of 32Question permalink

Analyzing +4 | -1

Statement-I: Leaching of gold with cyanide ion in absence of

air/{O_2}

leads to cyano complex of

Au\left( {III} \right).

Statement-II: Zinc is oxidized during the displacement reaction carried out for gold extraction.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: D

Statement-I is incorrect but statement-II is correct

Statement-I is incorrect because leaching of gold with cyanide in the presence of air/O2 leads to the formation of the

Au(I)

complex,

[Au(CN)_2]^-

. Statement-II is correct because zinc is oxidized to

Zn^{2+}

during the displacement reaction.

Solution approach

Correct Answer + Core Reason: Statement-I is incorrect but statement-II is correct. Leaching of gold forms $Au(I)$ complex, and zinc is oxidized in the displacement reaction.

Step-by-Step Solution:

Step 1: Analyze Statement-I: Leaching of gold with cyanide ion in the *presence* of air/O2 leads to the formation of the cyano complex of $Au(I)$ , not $Au(III)$ . The reaction is: $4Au + 8CN^- + O_2 + 2H_2O \rightarrow 4[Au(CN)_2]^- + 4OH^-$ . Therefore, Statement-I is incorrect.

Step 2: Analyze Statement-II: During the displacement reaction, zinc reduces the gold complex, and in the process, zinc itself is oxidized. The reaction is: $2[Au(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Au$ . Zinc goes from oxidation state 0 to +2, so it is oxidized. Therefore, Statement-II is correct.

Why other options are wrong:
- Option A: Statement-I is incorrect.
- Option B: Statement-I is incorrect.
- Option C: Statement-II is correct.

Key Concepts:
- Leaching of gold
- Displacement reaction
- Oxidation and reduction

Memory Aids:
- Gold forms a $Au(I)$ complex during leaching with cyanide.
- Zinc is oxidized during the displacement reaction.

Quick Tips:
- Remember the oxidation states of metals in their complexes.

4 Hard

4 of 32Question permalink

Understanding +4 | -1

Which of the following statements are incorrect?

I. Zinc can be extracted by self-reduction.

II. A depressant prevents certain type of particles to come to the froth.

III. Copper matte contains

{ZnS}

and

{Cu}_2{S}

.

IV. The solidified copper obtained from reverberatory furnace has blistered appearance due to evolution of

{SO}_2

during the extraction.

View answer and explanation

Correct option: C

I and III

Option C (I and III) is correct because Zinc cannot be extracted by self-reduction, and copper matte contains Cu2S and FeS, not ZnS.

Solution approach

Correct Answer + Core Reason: I and III. Zinc cannot be extracted by self-reduction and copper matte contains $Cu_2S$ and $FeS$ .

Step-by-Step Solution:

Step 1: Analyze each statement.

Step 2: I. Zinc cannot be extracted by self-reduction because the reduction potential of ZnO to Zn is too negative. It requires a stronger reducing agent like carbon.

Step 3: II. A depressant prevents certain types of particles from coming to the froth in froth flotation. This statement is correct.

Step 4: III. Copper matte contains primarily $Cu_2S$ and $FeS$ , not $ZnS$ .

Step 5: IV. The solidified copper obtained from the reverberatory furnace has a blistered appearance due to the evolution of $SO_2$ during the extraction. This statement is correct.

Why other options are wrong:
- Option A: Statement II is correct.
- Option B: Statement II is correct.
- Option D: Statement IV is correct.

Key Concepts:
- Self-reduction
- Froth flotation
- Copper matte composition

Memory Aids:
- Zinc needs carbon reduction, not self-reduction.

Quick Tips:
- Remember the composition of copper matte and the conditions for self-reduction.

5 Hard

5 of 32Question permalink

Analyzing +4 | -1

Which of the following statements about the advantages of roasting of sulphide ore before reduction is not true?

View answer and explanation

Correct option: D

Carbon and hydrogen are suitable reducing agents for metal sulphides

Option D is correct because carbon and hydrogen are generally not suitable reducing agents for metal sulfides directly; roasting converts the sulfide to an oxide, which is then reduced.

Solution approach

Correct Answer + Core Reason: Carbon and hydrogen are suitable reducing agents for metal sulphides. This statement is incorrect because metal sulfides are typically roasted to oxides before reduction with carbon or hydrogen.

Step-by-Step Solution:

Step 1: Understand the context: The question asks about the advantages of roasting sulfide ores before reduction.

Step 2: Analyze option A: $\Delta G_f^\circ$ of the sulfide is greater than $CS_2$ and $H_2S$ . This is generally true, making the formation of $CS_2$ or $H_2S$ less favorable than the metal sulfide itself.

Step 3: Analyze option B: $\Delta G_f^\circ$ is negative for roasting of sulfide ore to oxide. This is true, indicating that the roasting process is thermodynamically favorable.

Step 4: Analyze option C: Roasting of the sulfide to oxide is thermodynamically feasible. This is true, as indicated by the negative $\Delta G_f^\circ$ .

Step 5: Analyze option D: Carbon and hydrogen are suitable reducing agents for metal sulfides. This is generally false. Sulfides are first converted to oxides by roasting, and then the oxides are reduced by carbon or hydrogen.

Step 6: Conclusion: Option D is the incorrect statement.

Why other options are wrong:
- Option A: This statement is generally true. The Gibbs free energy of formation of metal sulfides is typically higher (more positive or less negative) than that of $CS_2$ and $H_2S$ , making the formation of $CS_2$ or $H_2S$ less favorable.
- Option B: This statement is true. Roasting sulfide ores to form oxides is a thermodynamically favorable process, indicated by a negative Gibbs free energy change.
- Option C: This statement is true. Because the Gibbs free energy change for roasting is negative, the process is thermodynamically feasible.
- Option D: This statement is incorrect. Metal sulfides are not directly reduced by carbon or hydrogen. They are first converted to oxides by roasting, and then the oxides are reduced.

Key Concepts:
- Thermodynamics of metallurgical processes
- Gibbs free energy of formation
- Roasting of sulfide ores
- Reduction of metal oxides

Memory Aids:
- Remember that sulfides are usually roasted to oxides before reduction.

Quick Tips:
- When dealing with metallurgical processes, consider the thermodynamic feasibility of each step.

6 Hard

6 of 32Question permalink

Remembering +4 | -1

In Mac Arthur forrest method, silver is extracted from the solution of

\mathrm{Na}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

by the use of

View answer and explanation

Correct option: D

Zn

Option D (Zn) is correct because Zinc is used as a reducing agent to precipitate silver from the complex ion solution in the Mac Arthur Forrest method.

Solution approach

Correct Answer + Core Reason: Zn. Zinc is used as a reducing agent to displace silver from the complex.

Step-by-Step Solution:

Step 1: Understand the Mac Arthur Forrest method. This method involves leaching silver ore with a cyanide solution to form a silver cyanide complex.

Step 2: The silver is then recovered from the solution by adding a more electropositive metal, which acts as a reducing agent.

Step 3: Zinc is more electropositive than silver and can displace it from the complex ion, $\mathrm{Na}[{\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}}]$ . The reaction is:
$\mathrm{2Na}[{\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}}] + \mathrm{Zn} \to \mathrm{Na}_2[{\mathrm{Zn}{{\left( {\mathrm{CN}} \right)}_4}}] + \mathrm{2Ag}$

Step 4: Silver precipitates out as a solid.

Why other options are wrong:
- Option A: Fe (Iron) is not typically used in the Mac Arthur Forrest method for silver extraction.
- Option B: Mg (Magnesium) is a strong reducing agent but not typically used in this specific process.
- Option C: Cu (Copper) is less electropositive than silver and cannot displace it from the cyanide complex.
- Option D: Zn (Zinc) is the correct reducing agent used in this process.

Key Concepts:
- Extraction of metals
- Redox reactions
- Electrochemistry
- Complex formation

Memory Aids:
- Zinc is more reactive than silver, so it can displace it.

Quick Tips:
- Remember the reactivity series to determine which metal can displace another.

7 Hard

7 of 32Question permalink

Understanding +4 | -1

Assertion: All the ores are mineral

Reason: Ores contains metals in combined state

View answer and explanation

Correct option: B

If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion

Option B is correct because while both the assertion and the reason are true, the reason does not explain why all ores are minerals; it explains why ores contain metals.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion. All ores are minerals, and ores contain metals in a combined state, but the latter doesn't explain the former.

Step-by-Step Solution:

Step 1: Analyze the assertion: All ores are minerals.
- This is true because ores are a specific type of mineral deposit that contains a high concentration of a valuable metal.

Step 2: Analyze the reason: Ores contain metals in a combined state.
- This is also true. Metals in ores are typically found as oxides, sulfides, carbonates, etc.

Step 3: Determine if the reason explains the assertion.
- The reason explains why we extract metals from ores, not why all ores are minerals.

Why other options are wrong:
- Option A: The reason is true, but it doesn't explain the assertion.
- Option C: Both the assertion and reason are true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Definition of ores and minerals
- Composition of ores

Memory Aids:
- All ores are minerals, but not all minerals are ores.

Quick Tips:
- Understand the definitions of key terms in metallurgy.

8 Hard

8 of 32Question permalink

Understanding +4 | -1

In the process of extraction of gold,

Roasted gold ore

+ \mathrm{C}{\mathrm{N}^ - }\mathrm{\;} + \mathrm{\;}{\mathrm{H}_2}\mathrm{O\;}\mathop \to \limits^{{\mathrm{O}_2}} \mathrm{\;}\left[ X \right]\mathrm{\;} + \mathrm{\;O}{\mathrm{H}^ - }

\left[ X \right]\mathrm{\;} + \mathrm{\;Zn\;} \to \mathrm{\;}\left[ Y \right]\mathrm{\;} + \mathrm{\;Au}

Identify the complexes [X] and [Y]

View answer and explanation

Correct option: A

X = {\left[ {\mathrm{Au}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - },\mathrm{\;}Y = {\left[ {\mathrm{Zn}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}

Option A is correct because it accurately represents the complexes formed during the leaching and subsequent zinc displacement in gold extraction.

Solution approach

Correct Answer + Core Reason: $X = {\left[ {\mathrm{Au}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - },\mathrm{\;}Y = {\left[ {\mathrm{Zn}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}$ . The gold complex formed is dicyanoaurate(I) and the zinc complex is tetracyanozincate(II).

Step-by-Step Solution:

Step 1: Identify the leaching process: Gold ore reacts with cyanide ions in the presence of oxygen and water to form a gold cyanide complex and hydroxide ions.

Step 2: Write the leaching reaction: $4Au + 8CN^- + O_2 + 2H_2O \rightarrow 4[Au(CN)_2]^- + 4OH^-$ \nSo, complex X is $[Au(CN)_2]^-$ .

Step 3: Identify the displacement reaction: The gold cyanide complex reacts with zinc to form a zinc cyanide complex and gold.

Step 4: Write the displacement reaction: $2[Au(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Au$ \nSo, complex Y is $[Zn(CN)_4]^{2-}$ .

Step 5: Combine the results: Complex X is $[Au(CN)_2]^-$ and complex Y is $[Zn(CN)_4]^{2-}$ .

Why other options are wrong:
- Option B: Incorrect because the gold complex formed during leaching is $[Au(CN)_2]^-$ not $[Au(CN)_4]^{3-}$ .
- Option C: Incorrect because the zinc complex formed during displacement is $[Zn(CN)_4]^{2-}$ not $[Zn(CN)_6]^{4-}$ .
- Option D: Incorrect because the gold complex formed during leaching is $[Au(CN)_2]^-$ not $[Au(CN)_4]^{-}$ .

Key Concepts:
- Leaching of gold
- Complex formation
- Displacement reaction

Memory Aids:
- Gold forms a dicyano complex during leaching.
- Zinc forms a tetracyano complex during displacement.

Quick Tips:
- Remember the common cyanide complexes of gold and zinc in extraction processes.

9 Hard

9 of 32Question permalink

Understanding +4 | -1

Assertion: In the smelting of copper ore coke is added in the blast furnace.

Reason: Coke reduces,

\mathrm{CuO}

into

\mathrm{Cu}

.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False

Option C (If Assertion is True but the Reason is False) is correct because coke is used in the smelting of copper ore, but its primary role is to produce heat, not directly reduce CuO to Cu.

Solution approach

Correct Answer + Core Reason: If Assertion is True but the Reason is False. Coke is added in the blast furnace during smelting of copper ore, but it is primarily used as a fuel to generate heat, not to directly reduce $CuO$ to $Cu$ .

Step-by-Step Solution:

Step 1: Analyze the assertion: Coke is indeed added to the blast furnace during the smelting of copper ore.

Step 2: Analyze the reason: While coke does act as a reducing agent in the blast furnace, its primary role in copper smelting is to provide heat for the process. The reduction of $CuO$ to $Cu$ is mainly achieved through self-reduction or by other reducing agents.

Step 3: Therefore, the assertion is true, but the reason is false.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Smelting of copper ore
- Role of coke in blast furnace
- Reducing agents

Memory Aids:
- Coke = Heat in copper smelting

Quick Tips:
- Remember the primary role of coke in copper smelting.

10 Hard

10 of 32Question permalink

Understanding +4 | -1

Match the extraction process listed in column I with metals listed in column II.

Column I Column II

A. Self reduction (P) Lead

B. Carbon reduction (Q) Silver

C. Complex formation and (R) Copper

displacement by metal

D. Decomposition of iodide (S) Boron

View answer and explanation

Correct option: B

\mathrm{A} - \mathrm{P},\mathrm{R};\mathrm{B} - \mathrm{P},\mathrm{R};\mathrm{C} - \mathrm{Q};\mathrm{D} - \mathrm{S}

Option B is correct because it correctly matches the extraction processes with the corresponding metals: Self-reduction (Pb, Cu), Carbon reduction (Pb, Cu), Complex formation and displacement (Ag), Decomposition of iodide (B).

Solution approach

Correct Answer + Core Reason: $\mathrm{A} - \mathrm{P},\mathrm{R};\mathrm{B} - \mathrm{P},\mathrm{R};\mathrm{C} - \mathrm{Q};\mathrm{D} - \mathrm{S}$ . This matching correctly pairs the extraction methods with the appropriate metals.

Step-by-Step Solution:

Step 1: Analyze each extraction process and the metals listed.

Step 2: Match the processes with the metals:
- Self-reduction: Lead (Pb) and Copper (Cu) (e.g., PbS, Cu2S)
- Carbon reduction: Lead (Pb) and Copper (Cu) (e.g., ZnO, Fe2O3)
- Complex formation and displacement by metal: Silver (Ag) (Cyanide process)
- Decomposition of iodide: Boron (B) (Van Arkel method)

Step 3: Combine the matches to form the correct option.

Why other options are wrong:
- Option A: Incorrect matching of carbon reduction and complex formation.
- Option C: Incorrect matching of carbon reduction and complex formation.
- Option D: Incorrect matching of self-reduction and carbon reduction.

Key Concepts:
- Metal extraction processes
- Self-reduction, Carbon reduction, Complex formation, Decomposition of iodide

Memory Aids:
- Self-reduction: Think of metals that can reduce themselves (e.g., Pb, Cu).
- Carbon reduction: Common method for many metals.
- Complex formation: Often used for precious metals like Ag.
- Decomposition of iodide: Van Arkel method for purification.

Quick Tips:
- Remember specific examples of metals extracted by each method.

11 Hard

11 of 32Question permalink

Understanding +4 | -1

Assertion : Zone refining is based on the fact that impurities are more soluble in molten state than in solid in the presence of oxygen.
Reason : This method is used to prepare pure metal oxides.

View answer and explanation

Correct option: D

If both Assertion & Reason are false

Option D (If both Assertion & Reason are false) is correct because zone refining relies on impurities being more soluble in the molten state *without* oxygen, and it's used for metals, not metal oxides.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are false. Zone refining is based on the fact that impurities are more soluble in molten state than in solid state, and it is used to obtain pure metals, not metal oxides.

Step-by-Step Solution:

Step 1: Analyze the assertion: Zone refining is based on the principle that impurities are more soluble in the molten state than in the solid state of a metal. The presence of oxygen is not a requirement.

Step 2: Analyze the reason: Zone refining is used to obtain ultra-pure metals, such as silicon and germanium, not metal oxides.

Step 3: Since both the assertion and the reason are incorrect, the correct answer is that both are false.

Why other options are wrong:
- Option A: Both Assertion and Reason are false.
- Option B: Both Assertion and Reason are false.
- Option C: Both Assertion and Reason are false.

Key Concepts:
- Zone refining
- Impurity solubility
- Purification of metals

Memory Aids:
- Zone refining = Pure Metals

Quick Tips:
- Remember the principle behind zone refining and its application.

12 Hard

12 of 32Question permalink

Application +4 | -1

Consider the following reaction occurring in the blast furnace.

\mathrm{Fe}_3 \mathrm{O}_{4(\mathrm{s})}+4 \mathrm{CO}_{(\mathrm{g})} \rightarrow 3 \mathrm{Fe}_{({l})}+4 \mathrm{CO}_{2(\mathrm{g})}

'x' kg of iron is produced when

2.32 \times 10^3 \mathrm{~kg}

\mathrm{Fe}_3 \mathrm{O}_4

and

2.8 \times 10^2 \mathrm{~kg}

\mathrm{CO}

are brought together in the furnace. The value of 'x' is ____. (nearest integer)

{Given:

Molar mass of

\mathrm{Fe}_3 \mathrm{O}_4=232 \mathrm{~g} \mathrm{~mol}^{-1}

Molar mass of

\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}

Molar mass of

\left.\mathrm{Fe}=56 \mathrm{~g} \mathrm{~mol}^{-1}\right\}

View answer and explanation

Correct option: A

420

Option A is correct because based on the stoichiometry of the reaction, 232 kg of Fe3O4 will produce 168 kg of Fe. Given the limiting reagent CO, 280 kg of CO will produce 420 kg of Fe.

Solution approach

Correct Answer + Core Reason: 420. The limiting reagent is CO, and based on the stoichiometry, 280 kg of CO will produce 420 kg of Fe.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$\mathrm{Fe}_3 \mathrm{O}_{4(\mathrm{s})}+4 \mathrm{CO}_{(\mathrm{g})} \rightarrow 3 \mathrm{Fe}_{({l})}+4 \mathrm{CO}_{2(\mathrm{g})}$

Step 2: Calculate the moles of $\mathrm{Fe}_3 \mathrm{O}_4$ and $\mathrm{CO}$ :
- Moles of $\mathrm{Fe}_3 \mathrm{O}_4 = \frac{2.32 \times 10^3 \text{ kg}}{232 \text{ g/mol}} = \frac{2.32 \times 10^6 \text{ g}}{232 \text{ g/mol}} = 10^4 \text{ mol}$
- Moles of $\mathrm{CO} = \frac{2.8 \times 10^2 \text{ kg}}{28 \text{ g/mol}} = \frac{2.8 \times 10^5 \text{ g}}{28 \text{ g/mol}} = 10^4 \text{ mol}$

Step 3: Determine the limiting reagent:
- From the balanced equation, 1 mole of $\mathrm{Fe}_3 \mathrm{O}_4$ reacts with 4 moles of $\mathrm{CO}$ .
- The mole ratio required is $\frac{\mathrm{CO}}{\mathrm{Fe}_3 \mathrm{O}_4} = 4$ .
- The actual mole ratio is $\frac{10^4}{10^4} = 1$ .
- Since the actual ratio is less than the required ratio, $\mathrm{CO}$ is the limiting reagent.

Step 4: Calculate the moles of Fe produced based on the limiting reagent:
- 4 moles of $\mathrm{CO}$ produce 3 moles of $\mathrm{Fe}$ .
- $10^4$ moles of $\mathrm{CO}$ produce $\frac{3}{4} \times 10^4$ moles of $\mathrm{Fe} = 7.5 \times 10^3$ moles of $\mathrm{Fe}$ .

Step 5: Calculate the mass of Fe produced:
- Mass of $\mathrm{Fe} = 7.5 \times 10^3 \text{ mol} \times 56 \text{ g/mol} = 42 \times 10^4 \text{ g} = 420 \text{ kg}$

Why other options are wrong:
- Option B: Incorrect calculation or wrong limiting reagent identification.
- Option C: Incorrect calculation or wrong limiting reagent identification.
- Option D: Incorrect calculation or wrong limiting reagent identification.

Key Concepts:
- Stoichiometry
- Limiting reagent
- Mole concept

Memory Aids:
- Remember to balance the chemical equation before performing calculations.

Quick Tips:
- Always identify the limiting reagent to determine the maximum amount of product formed.

13 Hard

13 of 32Question permalink

Remembering +4 | -1

Which of the following elements is considered as a metalloid?

View answer and explanation

Correct option: D

Te

Option D (

Te

) is correct because tellurium (Te) is generally considered a metalloid due to its intermediate properties between metals and nonmetals.

Solution approach

Correct Answer + Core Reason: Tellurium (Te) is a metalloid.

Step-by-Step Solution:

Step 1: Recall the definition of metalloids.

Step 2: Metalloids have properties intermediate between metals and non-metals.

Step 3: Tellurium (Te) is a well-known metalloid.

Why other options are wrong:
- Option A: Scandium (Sc) is a transition metal.
- Option B: Lead (Pb) is a metal.
- Option C: Bismuth (Bi) is a metal.

Key Concepts:
- Metalloids
- Properties of elements

Memory Aids:
- Remember the common metalloids: Boron, Silicon, Germanium, Arsenic, Antimony, Tellurium, Polonium.

Quick Tips:
- Know the common metalloids and their positions in the periodic table.

14 Hard

14 of 32Question permalink

Remembering +4 | -1

‘Lapis-Lazuli’ is a blue coloured precious stone. It is mineral of the class

View answer and explanation

Correct option: A

Sodium alumino silicate

Option A (Sodium alumino silicate) is correct because Lapis Lazuli is a metamorphic rock primarily composed of lazurite, a sodium aluminum silicate mineral.

Solution approach

Correct Answer + Core Reason: Sodium alumino silicate. Lapis-Lazuli is a sodium alumino silicate mineral.

Step-by-Step Solution:
Step 1: Recall the composition of Lapis-Lazuli.
Step 2: Lapis-Lazuli is primarily composed of lazurite, which is a sodium aluminum silicate with the formula ( $\mathrm{Na,Ca})_8(\mathrm{AlSiO}_4)_6(\mathrm{S,Cl,SO}_4)_2$ .

Why other options are wrong:
- Option B: Basic copper carbonate is the composition of Malachite, not Lapis-Lazuli.
- Option C: Zinc cobalt is not the primary composition of Lapis-Lazuli.
- Option D: Prussian blue is a pigment, not a mineral class.

Key Concepts:
- Mineral composition
- Silicate minerals

Memory Aids:
- Lapis-Lazuli contains sodium, aluminum, and silicon.

Quick Tips:
- Remember the composition of common minerals and gemstones.

15 Hard

15 of 32Question permalink

Remembering +4 | -1

Match the compound with the metal for which it is used for the process of extraction.

(i)

\mathrm{NaCN}

- (a) Titanium

(ii) Iodine - (b) Aluminium

(iii) Cryolite - (c) Silver ore

View answer and explanation

Correct option: A

(i) - (c), (ii)-(a), (iii)-(b)

Option A ((i) - (c), (ii)-(a), (iii)-(b)) is correct because NaCN is used for silver ore extraction, Iodine for Titanium, and Cryolite for Aluminium.

Solution approach

Correct Answer + Core Reason: (i) - (c), (ii)-(a), (iii)-(b). This matching correctly identifies the reagents used in the extraction of specific metals.

Step-by-Step Solution:

Step 1: NaCN is used in the cyanide process for the extraction of silver ore (Ag). The reaction involves the formation of a soluble silver cyanide complex.

Step 2: Iodine is used in the Van Arkel method for refining titanium (Ti). It forms a volatile titanium iodide that decomposes at high temperatures to yield pure titanium.

Step 3: Cryolite ( $Na_3AlF_6$ ) is used in the electrolytic extraction of aluminum (Al) to lower the melting point of alumina ( $Al_2O_3$ ) and increase its conductivity.

Why other options are wrong:
- Option B: Incorrectly matches Iodine with Aluminium and Cryolite with Titanium.
- Option C: Incorrectly matches NaCN with Titanium.
- Option D: Incorrectly matches NaCN with Aluminium.

Key Concepts:
- Extraction of metals
- Cyanide process
- Van Arkel method
- Electrolytic reduction

Memory Aids:
- NaCN with Silver, Iodine with Titanium, Cryolite with Aluminum.

Quick Tips:
- Remember the specific reagents used for the extraction of common metals.

16 Hard

16 of 32Question permalink

Understanding +4 | -1

Which statement is correct?

View answer and explanation

Correct option: D

Fluxes are carefully choosen to combine with the gangue present in the ore to produce easily fusible slag to carry away the impurities

Option D (Fluxes are carefully chosen to combine with the gangue present in the ore to produce easily fusible slag to carry away the impurities) is correct because fluxes are added to combine with gangue to form slag.

Solution approach

Correct Answer + Core Reason: Fluxes are carefully choosen to combine with the gangue present in the ore to produce easily fusible slag to carry away the impurities. Fluxes combine with gangue to form slag.

Step-by-Step Solution:

Step 1: Define the terms: gangue, flux, and slag.

Step 2: Gangue refers to the unwanted impurities present in the ore.

Step 3: Flux is a substance added to the ore to react with the gangue.

Step 4: Slag is the fusible product formed when the flux reacts with the gangue. It is easily removed from the molten metal.

Step 5: The purpose of adding a flux is to convert the gangue into a slag that can be easily separated.

Why other options are wrong:
- Option A: Slag does not combine with slag. Slag is the product, not a reactant.
- Option B: Gangues do not combine with slag. Gangues combine with flux.
- Option C: Gangues do not combine with flux to produce flux. They produce slag.

Key Concepts:
- Gangue, Flux, Slag definitions.
- Role of flux in metallurgy.

Memory Aids:
- Flux helps the impurities (gangue) to flow away as slag.

Quick Tips:
- Remember the relationship: Gangue + Flux → Slag.

17 Hard

17 of 32Question permalink

Remembering +4 | -1

Sperrylite is:

View answer and explanation

Correct option: B

\mathrm{PtA}{\mathrm{s}_2}

Option B (

\mathrm{PtA}{\mathrm{s}_2}

) is correct because Sperrylite is a platinum diarsenide mineral with the formula

\mathrm{PtA}{\mathrm{s}_2}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{PtA}{\mathrm{s}_2}$ . Sperrylite is platinum diarsenide.

Step-by-Step Solution:
Step 1: Recall the composition of Sperrylite.
Step 2: Sperrylite is a platinum arsenide mineral with the formula $\mathrm{PtA}{\mathrm{s}_2}$ .

Why other options are wrong:
- Option A: $\mathrm{AgCl}$ is silver chloride, also known as Horn Silver.
- Option C: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ is iron(III) oxide, also known as Hematite.
- Option D: $\mathrm{Sn}{\mathrm{O}_2}$ is tin(IV) oxide, also known as Cassiterite.

Key Concepts:
- Mineral composition
- Platinum minerals

Memory Aids:
- Sperrylite contains platinum and arsenic.

Quick Tips:
- Remember the composition of common minerals.

18 Hard

18 of 32Question permalink

Application +4 | -1

What is the EAN of

\mathrm{Ag}

in the product obtained after Leaching process?

View answer and explanation

Correct option: A

50

Option A (50) is correct because the EAN of Ag in the complex

[Ag(CN)_2]^-

is 47 (atomic number) - 1 (charge) + 4 (2 CN ligands * 2 electrons each) = 50.

Solution approach

Correct Answer + Core Reason: 50. The EAN (Effective Atomic Number) is calculated by adding the number of electrons on the metal ion and the number of electrons donated by the ligands.

Step-by-Step Solution:

Step 1: Identify the complex formed during leaching: In the leaching process, silver forms the complex $[Ag(CN)_2]^-$ .

Step 2: Determine the oxidation state of Ag: In $[Ag(CN)_2]^-$ , the oxidation state of Ag is +1.

Step 3: Calculate the number of electrons on the $Ag^+$ ion: The atomic number of Ag is 47, so $Ag^+$ has 47 - 1 = 46 electrons.

Step 4: Calculate the number of electrons donated by the ligands: Each $CN^-$ ligand donates 2 electrons, so 2 $CN^-$ ligands donate 2 * 2 = 4 electrons.

Step 5: Calculate the EAN: EAN = Number of electrons on the metal ion + Number of electrons donated by the ligands = 46 + 4 = 50.

Why other options are wrong:
- Option B: 0 is incorrect because it doesn't account for the electrons of the silver ion and the ligands.
- Option C: 0 is incorrect because it doesn't account for the electrons of the silver ion and the ligands.
- Option D: 0 is incorrect because it doesn't account for the electrons of the silver ion and the ligands.

Key Concepts:
- Effective Atomic Number (EAN)
- Coordination complexes
- Leaching process

Memory Aids:
- EAN = (Atomic number - oxidation state) + electrons from ligands

Quick Tips:
- Always identify the correct complex formed during the process before calculating EAN.

19 Hard

19 of 32Question permalink

Understanding +4 | -1

For the metallurgical process of which of the following calcined ore can be reduced by carbon?

View answer and explanation

Correct option: A

Haematite

Option A (Haematite) is correct because haematite (Fe2O3) is an iron oxide that can be reduced by carbon in a blast furnace.

Solution approach

Correct Answer + Core Reason: Haematite. Haematite (Fe2O3) is an iron ore that can be reduced by carbon.

Step-by-Step Solution:

Step 1: Identify the ores listed: Haematite, Chalcocite, Iron pyrites, and Sphalerite.

Step 2: Recall the composition of each ore:
- Haematite: Fe2O3
- Chalcocite: Cu2S
- Iron pyrites: FeS2
- Sphalerite: ZnS

Step 3: Consider which ores are typically reduced by carbon after calcination (or roasting).

Step 4: Oxides of iron, such as Fe2O3 (from haematite), can be reduced by carbon in a blast furnace.

Step 5: Sulphide ores like Cu2S, FeS2, and ZnS are typically first converted to oxides by roasting or calcination, and then the oxides can be reduced, but the question asks which calcined ore can be reduced by carbon directly.

Why other options are wrong:
- Option B: Chalcocite (Cu2S) is a sulfide ore and is typically roasted to form CuO before reduction.
- Option C: Iron pyrites (FeS2) is a sulfide ore and is typically roasted to form Fe2O3 before reduction.
- Option D: Sphalerite (ZnS) is a sulfide ore and is typically roasted to form ZnO before reduction.

Key Concepts:
- Reduction of metal oxides by carbon.
- Composition of common ores.
- Calcination and roasting processes.

Memory Aids:
- Remember that iron oxides are reduced by carbon in a blast furnace.

Quick Tips:
- Consider the chemical composition of the ore and whether it is an oxide or sulfide.

20 Hard

20 of 32Question permalink

Analyzing +4 | -1

Which series of reactions correctly represents chemical relations related to iron and its compound?

View answer and explanation

Correct option: D

Fe\xrightarrow{{{O_2},\,\,Heat}}F{e_3}{O_4}\xrightarrow{{CO,\,\,600^\circ \,C}}\,FeO\xrightarrow{{CO,\,\,700^\circ \,C}}Fe\,

Option D is correct because it accurately depicts the stepwise reduction of iron oxide (

Fe_3O_4

) to iron (Fe) using CO at different temperatures.

Solution approach

Correct Answer + Core Reason: $Fe\xrightarrow{{{O_2},\,\,Heat}}F{e_3}{O_4}\xrightarrow{{CO,\,\,600^\circ \,C}}\,FeO\xrightarrow{{CO,\,\,700^\circ \,C}}Fe\,$ . This series correctly represents the reactions of iron and its compounds.

Step-by-Step Solution:

Step 1: Iron reacts with oxygen upon heating to form iron(II,III) oxide ( $Fe_3O_4$ ).
$3Fe + 2O_2 \xrightarrow{Heat} Fe_3O_4$

Step 2: $Fe_3O_4$ is reduced to iron(II) oxide (FeO) by carbon monoxide at 600°C.
$Fe_3O_4 + CO \xrightarrow{600^\circ C} 3FeO + CO_2$

Step 3: Iron(II) oxide (FeO) is further reduced to iron (Fe) by carbon monoxide at 700°C.
$FeO + CO \xrightarrow{700^\circ C} Fe + CO_2$

Why other options are wrong:
- Option A: The direct conversion of $FeSO_4$ to $Fe_2(SO_4)_3$ requires an oxidizing agent. Heating $Fe_2(SO_4)_3$ will not directly yield Fe.
- Option B: Iron reacts with oxygen to form $Fe_3O_4$ , not $FeO$ directly. Also, heating $FeSO_4$ does not directly produce Fe.
- Option C: $FeCl_3$ does not directly convert to $FeCl_2$ upon heating in air. The conversion of $FeCl_2$ to Fe requires a stronger reducing agent than Zn if water is present.

Key Concepts:
- Redox reactions of iron
- Formation of iron oxides
- Reduction of iron oxides using CO

Memory Aids:
- Remember that iron can form multiple oxides ( $FeO$ , $Fe_2O_3$ , $Fe_3O_4$ ).
- CO acts as a reducing agent at high temperatures.

Quick Tips:
- Pay attention to the oxidation states of iron in different compounds.
- Consider the reaction conditions (temperature, presence of oxygen/reducing agents) when predicting the products.

21 Hard

21 of 32Question permalink

Application +4 | -1

The overall reaction for the electrolytic production of aluminium by means of the Hall Heroult process may be represented as

\mathrm{Al_2O}_3(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \longrightarrow 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g})

At

1000^{\circ} \mathrm{C}

, the standard free energy change for the process is

-594 \mathrm{~kJ} \mathrm{~mol}^{-1}

. The minimum voltage required to produce one mole aluminium at this temperature is

View answer and explanation

Correct option: A

1.02 \mathrm{~V}

Option A (1.02 V) is correct because it is calculated using the formula

\Delta G = -nFE

, where n is the number of moles of electrons transferred, F is Faraday's constant, and E is the cell potential.

Solution approach

Correct Answer + Core Reason: $1.02 \mathrm{~V}$ . The minimum voltage is calculated using the Gibbs free energy change and Faraday's constant.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta G = -594 \mathrm{~kJ} \mathrm{~mol}^{-1} = -594000 \mathrm{~J} \mathrm{~mol}^{-1}$ . The reaction is $\mathrm{Al_2O}_3(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \longrightarrow 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g})$ .

Step 2: Determine the number of moles of electrons transferred (n): For 2 moles of Al, the number of electrons transferred is $2 \times 3 = 6$ . So, $n = 6$ .

Step 3: Use the formula: $\Delta G = -nFE$ , where $F = 96500 \mathrm{~C} \mathrm{~mol}^{-1}$ (Faraday's constant).

Step 4: Rearrange the formula to solve for E: $E = -\frac{\Delta G}{nF}$

Step 5: Substitute the values: $E = -\frac{-594000 \mathrm{~J} \mathrm{~mol}^{-1}}{6 \times 96500 \mathrm{~C} \mathrm{~mol}^{-1}}$

Step 6: Calculate the voltage: $E = \frac{594000}{6 \times 96500} = \frac{594000}{579000} \approx 1.02 \mathrm{~V}$

Why other options are wrong:
- Option B: 2.05 V is incorrect. It might be obtained by not considering the 2 moles of Al produced or by using an incorrect value for 'n'.
- Option C: 3.08 V is incorrect. This value doesn't correspond to the correct application of the formula and the given data.
- Option D: 4.03 V is incorrect. This value is significantly different from the correct voltage and arises from an incorrect calculation.

Key Concepts:
- Gibbs free energy
- Electrolysis
- Faraday's constant
- Cell potential

Memory Aids:
- $\Delta G = -nFE$ (Gibbs free energy and cell potential relationship)

Quick Tips:
- Ensure the units are consistent before substituting into the formula.
- Correctly determine the number of moles of electrons transferred.

22 Hard

22 of 32Question permalink

Remembering +4 | -1

An ore of silver is:

View answer and explanation

Correct option: A

Argentite

Option A (Argentite) is the correct answer because Argentite (

\mathrm{Ag}_2\mathrm{S}

) is a sulfide mineral and an important ore of silver.

Solution approach

Correct Answer + Core Reason: Argentite. Argentite ( $\mathrm{Ag}_2\mathrm{S}$ ) is an ore of silver.

Step-by-Step Solution:
Step 1: Identify the chemical formulas of the given options.
Step 2: Recall that Argentite ( $\mathrm{Ag}_2\mathrm{S}$ ) is a common ore of silver.

Why other options are wrong:
- Option B: Stibnite ( $\mathrm{Sb}_2\mathrm{S}_3$ ) is an ore of antimony, not silver.
- Option C: Haematite ( $\mathrm{Fe}_2\mathrm{O}_3$ ) is an ore of iron, not silver.
- Option D: Bauxite ( $\mathrm{Al}_2\mathrm{O}_3.\mathrm{xH}_2\mathrm{O}$ ) is an ore of aluminum, not silver.

Key Concepts:
- Ores of silver, antimony, iron, and aluminum.

Memory Aids:
- Argentite contains silver.

Quick Tips:
- Remember the common ores of important metals.

23 Hard

23 of 32Question permalink

Understanding +4 | -1

Consider the following statements-

(A) In theAluminothermite process, aluminium acts as reducing agent.

(B) The process of extraction of gold involves the formation of

\left[\mathrm{Au}(\mathrm{CN})_2\right]^{-}

and

\left[\mathrm{Zn}(\mathrm{CN})_4\right]^{2-}

.
(C) In the extractive metallurgy of zinc, partial fusion of

\mathrm{ZnO}

with coke is called sintering and reduction of ore to the molten metal is called smelting.

(D) Extractive metallurgy of silver from its ore argentine involves complex formation and displacement by more electropositive metal.

Choose the correct options-

View answer and explanation

Correct option: D

A, B, C and D

Option D is correct because all the statements (A, B, C, and D) are factually accurate regarding the extraction of metals.

Solution approach

Correct Answer + Core Reason: A, B, C and D. All the statements are correct.

Step-by-Step Solution:

Step 1: Analyze each statement individually.

Step 2: Verify the correctness of each statement:
- (A) Aluminothermite process: Aluminum acts as a reducing agent (correct).
- (B) Gold extraction: Formation of [Au(CN)2]- and [Zn(CN)4]2- (correct).
- (C) Zinc metallurgy: Sintering and smelting definitions (correct).
- (D) Silver metallurgy: Complex formation and displacement (correct).

Step 3: Since all statements are correct, choose the option that includes all of them.

Why other options are wrong:
- Option A: Missing statements C and D.
- Option B: Missing statements A and D.
- Option C: Missing statement D.

Key Concepts:
- Aluminothermite process
- Gold extraction (cyanide process)
- Zinc metallurgy (sintering, smelting)
- Silver metallurgy

Memory Aids:
- Remember the specific reagents and processes involved in each extraction.

Quick Tips:
- Review the standard extraction methods for common metals.

24 Hard

24 of 32Question permalink

Analyzing +4 | -1

Assertion (A): In the extraction of

{Ag}

by Cyanide process, Air must be blown in the disolution of

{Ag}_2 {S}

in dil.

{NaCN}

Reason (R): Continuous air blow oxidises

{Na}_2{S}

to

{Na}_2 {SO}_3

and prevents reaction between

{Na}_3{S}

and

{Na}\left[{Ag}({CN})_2\right]

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

Option A is correct because in the cyanide process for silver extraction, air is blown to oxidize Na2S to Na2SO3, preventing its reaction with Na[Ag(CN)2].

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true and (R) is the correct explanation of (A). Air is blown to oxidize Na2S to Na2SO3, preventing its reaction with Na[Ag(CN)2].

Step-by-Step Solution:

Step 1: Analyze the assertion: In the extraction of Ag by the Cyanide process, Air must be blown in the dissolution of Ag2S in dil. NaCN.
- This is true. Air is necessary for the reaction to proceed efficiently.

Step 2: Analyze the reason: Continuous air blow oxidises Na2S to Na2SO3 and prevents reaction between Na3S and Na[Ag(CN)2].
- This is also true. The oxidation of Na2S prevents it from interfering with the silver complex.

Step 3: Determine if the reason explains the assertion.
- The reason correctly explains why air is blown into the solution.

Why other options are wrong:
- Option B: The reason is the correct explanation.
- Option C: Both are true.
- Option D: Both are true.

Key Concepts:
- Cyanide process for silver extraction
- Role of air in the process

Memory Aids:
- Air prevents sulfide interference in silver extraction.

Quick Tips:
- Understand the purpose of each step in the cyanide process.

25 Hard

25 of 32Question permalink

Remembering +4 | -1

Which is not employed for refining of metal?

View answer and explanation

Correct option: B

Leaching

Option B (Leaching) is correct because leaching is an extraction method, not a refining method.

Solution approach

Correct Answer + Core Reason: Leaching. Leaching is an extraction process, not a refining process.

Step-by-Step Solution:

Step 1: Define Refining: Refining is the process of purifying a metal from its impurities.

Step 2: Define Poling: Poling is a refining method used for copper and tin, where green wood is used to reduce metal oxides.

Step 3: Define Leaching: Leaching is a process of extracting a soluble substance from a solid by dissolving it in a liquid. This is an extraction method, not a refining method.

Step 4: Define Electrolysis: Electrolysis is a refining method where an electric current is used to purify a metal.

Step 5: Define Liquation: Liquation is a refining method where a metal is separated from impurities based on differences in melting points.

Why other options are wrong:
- Option A: Poling is a refining method.
- Option C: Electrolysis is a refining method.
- Option D: Liquation is a refining method.

Key Concepts:
- Refining Methods
- Extraction Methods
- Leaching
- Poling
- Electrolysis
- Liquation

Memory Aids:
- Leaching: Extraction, not refining

Quick Tips:
- Distinguish between extraction and refining processes.
- Know the purpose of each metallurgical process.

26 Hard

26 of 32Question permalink

Application +4 | -1

The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by the hall process is
(Atomic mass of Al=27)

View answer and explanation

Correct option: D

90 kg

Option D (90 kg) is correct because the balanced equation shows that 4 moles of Al are produced for every 3 moles of C consumed, and the calculation confirms this.

Solution approach

Correct Answer + Core Reason: 90 kg. The stoichiometry of the reaction dictates the mass of carbon consumed.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the Hall-Héroult process. $2\mathrm{Al}_2\mathrm{O}_3 + 3\mathrm{C} \rightarrow 4\mathrm{Al} + 3\mathrm{CO}_2$

Step 2: Calculate the moles of aluminum produced. Moles of Al = mass / atomic mass = 270 kg / 27 g/mol = 10000 moles.

Step 3: Determine the moles of carbon consumed. From the balanced equation, 4 moles of Al are produced for every 3 moles of C consumed. Therefore, moles of C = (3/4) * moles of Al = (3/4) * 10000 = 7500 moles.

Step 4: Calculate the mass of carbon consumed. Mass of C = moles * atomic mass = 7500 moles * 12 g/mol = 90000 g = 90 kg.

Why other options are wrong:
- Option A: 180kg. This is incorrect. This value might be obtained by incorrectly assuming a 1:1 mole ratio between Al and C, or by not properly accounting for the stoichiometry of the reaction.
- Option B: 270 kg. This is incorrect. This value might be obtained by assuming that the mass of carbon consumed is equal to the mass of aluminum produced, which is not true due to the stoichiometry of the reaction.
- Option C: 540 kg. This is incorrect. This value is significantly higher than the correct answer and likely results from a calculation error or misunderstanding of the stoichiometry.

Key Concepts:
- Hall-Héroult Process
- Stoichiometry
- Moles
- Atomic Mass

Memory Aids:
- Remember the balanced equation for the Hall-Héroult process.

Quick Tips:
- Always start with a balanced chemical equation to determine the correct mole ratios.

27 Hard

27 of 32Question permalink

Remembering +4 | -1

Correct statement is

View answer and explanation

Correct option: A

van-Arkel method is used for extraction of Zr

Option A (van-Arkel method is used for extraction of Zr) is correct because the van-Arkel method is an iodide refining process used for the purification of metals like zirconium and titanium.

Solution approach

Correct Answer + Core Reason: van-Arkel method is used for Zr extraction.

Step-by-Step Solution:

Step 1: Recall the van-Arkel method.

Step 2: The van-Arkel method is used for the purification of metals like Zr and Ti.

Why other options are wrong:
- Option B: Limestone is a basic flux, not acidic.
- Option C: Dolomite is an ore of magnesium and calcium, not aluminum.
- Option D: Willemite is a silicate ore, not a carbonate ore.

Key Concepts:
- Metal extraction methods
- Fluxes
- Ores of metals

Memory Aids:
- van-Arkel = Zirconium (association).

Quick Tips:
- Know the specific methods used for extracting and refining different metals.

28 Hard

28 of 32Question permalink

Understanding +4 | -1

An Ellingham diagram provides information about

View answer and explanation

Correct option: C

the temperature dependence of the standard Gibbs energies of formation of some metal oxides.

Option C (the temperature dependence of the standard Gibbs energies of formation of some metal oxides) is correct because Ellingham diagrams plot the change in Gibbs free energy of formation of metal oxides versus temperature.

Solution approach

Correct Answer + Core Reason: the temperature dependence of the standard Gibbs energies of formation of some metal oxides. Ellingham diagrams plot Gibbs free energy vs temperature.

Step-by-Step Solution:

Step 1: Define the Ellingham diagram.\nAn Ellingham diagram is a graph plotting the standard Gibbs free energy change ( $\Delta G^\circ$ ) for the formation of metal oxides as a function of temperature.

Step 2: Understand the information provided by the diagram.\nThe diagram provides information about the thermodynamic stability of metal oxides at different temperatures and can be used to predict the feasibility of reducing a metal oxide with a given reducing agent.

Step 3: Analyze the given options.
- Option A is incorrect because Ellingham diagrams do not directly explain the kinetics of the reduction process.
- Option B is incorrect because Ellingham diagrams do not directly show the pressure dependence of standard electrode potentials.
- Option D is incorrect because Ellingham diagrams do not show the conditions of pH and potential under which a species is thermodynamically stable (Pourbaix diagrams do that).

Why other options are wrong:
- Option A: Ellingham diagrams provide thermodynamic information, not kinetic information.
- Option B: Ellingham diagrams do not directly show the pressure dependence of electrode potentials.
- Option D: Ellingham diagrams do not show pH and potential stability conditions.

Key Concepts:
- Ellingham Diagram: Gibbs free energy vs. temperature for metal oxide formation.
- Gibbs Free Energy: A thermodynamic potential that measures the amount of energy available to do work.

Memory Aids:
- Ellingham: "Energy" and "Temperature".

Quick Tips:
- Remember that Ellingham diagrams are used to assess the thermodynamic feasibility of metal oxide reduction.

29 Hard

29 of 32Question permalink

Remembering +4 | -1

Schweitzer’s reagent used for dissolving cellulose in the manufacture of artificial silk is:

View answer and explanation

Correct option: C

\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_4}\mathrm{S}{\mathrm{O}_4}

Option C (

\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_4}\mathrm{S}{\mathrm{O}_4}

) is correct because Schweitzer's reagent is a solution of tetraamminecopper(II) hydroxide, which is formed by dissolving copper(II) hydroxide in ammonia.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_4}\mathrm{S}{\mathrm{O}_4}$ . Schweitzer’s reagent is tetraamminecopper(II) sulfate.

Step-by-Step Solution:
Step 1: Recall the composition of Schweitzer's reagent.
Step 2: Schweitzer's reagent is a complex of copper(II) ions with ammonia, specifically $\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_4}^{2+}$ ions.
Step 3: Identify the correct formula from the options.

Why other options are wrong:
- Option A: $\mathrm{CuS}{\mathrm{O}_4}.5{\mathrm{H}_2}\mathrm{O}$ is copper(II) sulfate pentahydrate, also known as blue vitriol, and is not Schweitzer's reagent.
- Option B: CuI is copper(I) iodide, and is not Schweitzer's reagent.
- Option D: $\mathrm{Cu}{\left( {\mathrm{C}{\mathrm{H}_3}\mathrm{COO}} \right)_2}.\mathrm{Cu}{\left( {\mathrm{OH}} \right)_2}$ is basic copper acetate, and is not Schweitzer's reagent.

Key Concepts:
- Coordination complexes
- Schweitzer's reagent

Memory Aids:
- Schweitzer's reagent contains copper and ammonia.

Quick Tips:
- Remember the composition of important reagents.

30 Hard

30 of 32Question permalink

Understanding +4 | -1

The temperature of blast furnace to produce iron from its ore

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}

varies from 500

^\circ C

at the top of the furnace to about 1900

^\circ C

at the bottom of the furnace. The reaction between the ore

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}

and CO at the lowest temperature (

\sim

500

^\circ C

) is:

View answer and explanation

Correct option: C

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} + 3\mathrm{CO} \to 2\mathrm{Fe} + 3\mathrm{C}{\mathrm{O}_2}

The correct answer is C because at the lowest temperature in the blast furnace, iron(III) oxide is directly reduced to iron by carbon monoxide.

Solution approach

Correct Answer + Core Reason: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} + 3\mathrm{CO} \to 2\mathrm{Fe} + 3\mathrm{C}{\mathrm{O}_2}$ . At lower temperatures, the reduction of iron oxide to iron is most efficient with CO.

Step-by-Step Solution:

Step 1: Understand the process: In a blast furnace, iron ore ( $Fe_2O_3$ ) is reduced to iron using carbon monoxide (CO) as the reducing agent.

Step 2: Consider the temperature: At the lowest temperature zone ( $\sim 500^{\circ}C$ ), the primary reaction involves the direct reduction of $Fe_2O_3$ to Fe by CO.

Step 3: Write the balanced chemical equation: $Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)$

Why other options are wrong:
- Option A: This reaction represents the formation of $Fe_3O_4$ , which occurs at slightly higher temperatures than the direct reduction to Fe.
- Option B: This reaction represents the formation of $FeO$ , which is an intermediate step in the reduction process, but not the final product at the lowest temperature.
- Option D: This reaction is not balanced and does not accurately represent the reduction process.

Key Concepts:
- Blast furnace process
- Reduction of iron ore
- Role of carbon monoxide

Memory Aids:
- CO is the primary reducing agent in the blast furnace.
- Lower temperatures favor direct reduction to Fe.

Quick Tips:
- Remember the temperature zones and corresponding reactions in the blast furnace.

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