Question

$H_2{O}_2$ oxidizes $S{n}^{2+}$ , the products being $S{n}^{4+}$ and water. Remaining $H_2{O}_2$ decomposes slowly at room temperature to yield $O_2$ and water.
$200gm$ of 10% by mass $H_2{O}_2$ in water is treated with $100ml2MS{n}^{2+}$ and then the mixture is allowed to stand until no further reaction occurs. The volume of $O_2$ produced is: { $O_2$ is produced at $20oCand1atm$ }

• A. 4.669 L
• B. 5.221 L
• C. 6.113 L
• D. 4.442 L

Answer 1

Answer: (A)

Given Eq. of $H_2{O}_2$ = $\frac{10\%of200g}{17}$ = 1.1765
Eq. of $H_2{O}_2$ reduced completely = Eq of Sn²⁺
= 0.1 × (2 × 2)
= 0.4
$\therefore$ Remaining equivalents of $H_2{O}_2$
= 1.1764 - 0.4 = 0.7765
Or remaining moles of $H_2{O}_2$ $\frac{0.7765}{2}$ = 0.3882
there moles of $H_2{O}_2$ decomposes to yield $O_2\&H_{2}O$ as
$H_2{O}_2\to H_{2}O+\frac{1}{2}{O}_2$
$\therefore$ 1 1/2
$\therefore$ 0.3882 0.3882/2 = 0.1941
Now from V = $\frac{nRT}{P}$
The vol. of O₂ produced
V = $\frac{0.1941\times 0.0821\times 293}{1}$
V = 4.669 L