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Q. How many mL of 22.4 volume $H_{2}O_{2}$ is required to oxidise 0.1 mol of $H_{2}S$ gas to S?

NTA AbhyasNTA Abhyas 2020Some Basic Concepts of Chemistry

Solution:

$\underset{1 \text{mol}}{H_{2} O_{2}}+\underset{1 \text{mol}}{H_{2} S} \rightarrow 2H_{2}O+S$

0.1 mol $H_{2}S$ require $H_{2}O_{2}=0.1$ mol

Molarity of 22.4 volume $H_{2}O_{2}=\frac{22.4}{11.2}=2\text{M}$

Now volume of $H_{2}O_{2}$ required

$=\frac{m o l}{m o l a r i t y}=\frac{0.1}{2}L$

$= \frac{0.1 \, \times \, 1000}{2} \, = 50 \, \text{mL}$