Question

$H_{2}O_{2}$ oxidizes $Sn^{2 +}$ , the products being $Sn^{4 +}$ 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 $100ml2MSn^{2 +}$ 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 $H_{2}O_{2}$ = $\frac{10 \% \, o f \, 200 g}{17}$ = 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} \rightarrow H_{2}O+\frac{1}{2} \, O_{2}$
$\therefore $ 1 1/2
$\therefore $ 0.3882 0.3882/2 = 0.1941
Now from V = $\frac{n R T}{P}$
The vol. of O₂ produced
V = $\frac{0.1941 \times 0.0821 \times 293}{1}$
V = 4.669 L