In Kipp’s apparatus, ${{\rm{H}}_2}{\rm{S}}$ is prepared:

${\rm{By\;FeS}} + {\rm{dil}}.{\rm{\;}}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${\rm{By\;FeS}} + {\rm{conc}}.{\rm{\;}}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${\rm{By\;Fe}} + {\rm{dil}}.{\rm{\;}}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

JEE CHEMISTRY MCQ: P-BLOCK ELEMENTS

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In Kipp’s apparatus,

{\mathrm{H}_2}\mathrm{S}

is prepared:

View answer and explanation

Correct option: C

\mathrm{By\;FeS} + \mathrm{dil}.\mathrm{\;}{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

Option C (

\mathrm{By\;FeS} + \mathrm{dil}.\mathrm{\;}{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

) is correct because Kipp's apparatus uses the reaction between ferrous sulfide (FeS) and dilute sulfuric acid to produce H2S gas.

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17^{\text {th }}

{O}_2

{O}_2

\mathrm{C}{\mathrm{N}^ - }

{\mathrm{N}_2}

\mathrm{C}{\mathrm{N}^ - },\mathrm{\;}{\mathrm{N}_2}

2 {Fe}^{3+}+{SO}_2+2 {H}_2 {O} \rightarrow 2 {Fe}^{2+}+{SO}_4^{2-}+4 {H}^{+}

{H}_2 {~S}_2 {O}_8

{S}={O}

{S}-{OH}

{O}-

{O}

{NH}_3

{H}_2 {SO}_4

{H}_2 {SO}_4