Question

Two samples of potassium permanganate and mercuric oxide were heated separately. The amount of oxygen formed was in 3:4 ratio. The entire oxygen obtained is made to react with a lower oxide of sulphur formed by burning 12.8 g of sulphur. The volume of product formed at STP is:

• A. $8.96$ litre
• B. $11.2$ litre
• C. $22.4$ litre
• D. $44.8$ litre

Answer 1

The correct option is A $8.96$ litre

The reactions are as follows:
$2KMn{O}_4\stackrel{\Delta }{\to }{K}_{2}Mn{O}_4+Mn{O}_2+O_2$ $........\left(i\right)$
$2HgO\stackrel{\Delta }{\to }2Hg+O_2$ $........\left(ii\right)$
$S+O_2⟶S{O}_2$

$2S{O}_2+O_2⟶2S{O}_3$
$32$ g sulphur $⟶64$ g $S{O}_2$
$12.8$ g sulphur $⟶\frac{64\times 12.8}{32}$ $=25.6$ g $S{O}_2$
$128$ g $S{O}_2$ requires $1$ mole $O_2$
$25.6$ g $S{O}_2$ requires $=\frac{25.6\times 1}{128}$ moles $=0.2$ moles $O_2=6.4$ g $O_2$
Ratio of $O_2$ formed in reactions $\left(i\right)$ and $\left(ii\right)$ is $3:4$.
Amount of $O_2$ formed from $KMn{O}_4=\frac{6.4}{7}\times 3=2.74$ g $=0.08$ moles
$1$ mole $O_2$ is produced by $⟶2$ moles $KMn{O}_4$
$0.08$ mole $O_2⟶0.16KMn{O}_4$
$0.16$ moles $KMn{O}_4=25.28$ g
$1$ mole $O_2⟶2$ moles $HgO$
$0.0085$ mole $O_2⟶2\times 0.0085=0.017$ moles
$0.017$ moles $HgO⟶3.66$ g
$25.6$ g $S{O}_2=0.4$ moles
$2$ moles $S{O}_2⟶2$ moles $S{O}_3$
$0.4$ moles $S{O}_2⟶0.4$ moles $S{O}_3$
Volume of $S{O}_3$ at STP $=0.4\times 22.4$ $=8.96l$