Chemistry · 2021
JEE Main · 3 August 2021, Shift 2 · Q51
Chlorine is prepared according to the following equation: 4 HCl + MnO_2( s ) → 2 H_2 O_(l)+ MnCl_2( aq ) + Cl_2( g ). 10 g sample of MnO_2 produces…
Chlorine is prepared according to the following equation :
$$4 \mathrm{HCl}+\mathrm{MnO}_{2(\mathrm{~s})} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}_{(l)}+\mathrm{MnCl}_{2(\mathrm{aq})}+\mathrm{Cl}_{2(\mathrm{~g})} .
$$
$\displaystyle 10$ g sample of $\displaystyle \mathrm{MnO}_2$ produces $\displaystyle 2.24$ L of chlorine under SATP, the percentage purity of the $\displaystyle \mathrm{MnO}_2$ sample is \_\_\_\_. (Nearest integer)
[Atomic weight $\displaystyle =\mathrm{H}: 1.0, \mathrm{O}: 16.0, \mathrm{C}: 12.0, \mathrm{Cl}: 35.5, \mathrm{Mn}: 55.0$ ]
$\displaystyle \left[\mathrm{SATP}: \mathrm{T}=298 \mathrm{~K}, \mathrm{P}=10^5 \mathrm{~Pa}\right]$
Official answer
87
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.