Question

One litre of acidified $KMn{O}_4$ containing 15.8 g $KMn{O}_4$ (mol.wt=158) is decolourised by passing sufficient $S{O}_2$. If $S{O}_2$ is produced by $Fe{S}_2$, then which of the following options is/are correct?
$(Fe=56,S=32)$

• A. Mole of $Fe{S}_2$ consumed in this reaction are 0.250
• B. $S{O}_2$ produced from $Fe{S}_2$ will occupy 5.6 lit. at STP
• C. Equivalent weight of $KMn{O}_4$ in this reaction is $31.6g$
• D. Equivalent weight of $Fe{S}_2$ in this reaction is $10.9g$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $S{O}_2$ produced from $Fe{S}_2$ will occupy 5.6 lit. at STP
C Equivalent weight of $KMn{O}_4$ in this reaction is $31.6g$
D Equivalent weight of $Fe{S}_2$ in this reaction is $10.9g$

Since $S{O}_2$ is produced from $Fe{S}_2$, so
$Fe{S}_2+Mn{O}_4^{-}\stackrel{H^{+}}{\to }F{e}^{3+}+2S{O}_2+M{n}^{2+}$
Here n-factor of $Fe{S}_2$=11,
n-factor of $KMn{O}_4$=5
So, equivalent weight of $KMn{O}_4=\frac{M}{5}=\frac{158}{5}=31.6g$
Equivalent weight of $Fe{S}_2=\frac{M}{11}=\frac{120}{11}=10.9g$
Also, $KMn{O}_4+S{O}_2\stackrel{H^{+}}{\to }M{n}^{2+}+S{O}_4^{2-}$
Again $gmeq.ofKMn{O}_4=gmeq.ofS{O}_2$
$\Rightarrow 0.1\times 5=molesofS{O}_2\times 2$
$\therefore$ moles of $S{O}_2$ = 0.25 mol
So volume of $S{O}_2$ at STP=$0.25\times 22.4=5.6L$
Again, moles of $Fe{S}_2$ used = $\frac{1}{2}$ moles of $S{O}_2$ produced.
$\therefore$ moles of $Fe{S}_2$ = $\frac{1}{2}\times 0.25=0.125mol$