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Calculate the approximate mass (in g) of $\mathrm{H}_2 \mathrm{~S}$ required for the following reaction when $15 \mathrm{~L}$ of oxygen at STP reacts completely.
$x \mathrm{H}_2 \mathrm{~S}(g)+y \mathrm{O}_2(g) \longrightarrow a \mathrm{SO}_2(g)+b \mathrm{H}_2 \mathrm{O}(g)$
( molar mass of $\mathrm{H}_2 \mathrm{~S}=34 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Options:
$x \mathrm{H}_2 \mathrm{~S}(g)+y \mathrm{O}_2(g) \longrightarrow a \mathrm{SO}_2(g)+b \mathrm{H}_2 \mathrm{O}(g)$
( molar mass of $\mathrm{H}_2 \mathrm{~S}=34 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Solution:
2233 Upvotes
Verified Answer
The correct answer is:
15.16
$2 \mathrm{H}_2 \mathrm{~S}+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$
From balanced equation
3 moles of $\mathrm{O}_2$ reacts with 2 moles of $\mathrm{H}_2 \mathrm{~S}$
$22.4 \mathrm{~L}$ of $\mathrm{O}_2=1 \mathrm{moI}$
$$
\therefore 15 \mathrm{~L} \text { of } \mathrm{O}_2=\left(\frac{1}{22.4} \times 15\right) \mathrm{moI}
$$
3 moles of oxygen uses 2 moles of $\mathrm{H}_2 \mathrm{~S}$.
$\therefore\left(\frac{15}{224}\right)$ moles of $\mathrm{O}_2$ uses $\left(\frac{2}{3} \times \frac{15}{22.4}\right)$ moles of $\mathrm{H}_2 \mathrm{~S}$
Molar mass of $\mathrm{H}_2 \mathrm{~S}=34 \mathrm{~g} / \mathrm{moI}$
$\therefore$ Gram of $\mathrm{H}_2 \mathrm{~S}=\left(34 \times \frac{2}{3} \times \frac{15}{224}\right) \mathrm{g}=15.17 \mathrm{~g}$
From balanced equation
3 moles of $\mathrm{O}_2$ reacts with 2 moles of $\mathrm{H}_2 \mathrm{~S}$
$22.4 \mathrm{~L}$ of $\mathrm{O}_2=1 \mathrm{moI}$
$$
\therefore 15 \mathrm{~L} \text { of } \mathrm{O}_2=\left(\frac{1}{22.4} \times 15\right) \mathrm{moI}
$$
3 moles of oxygen uses 2 moles of $\mathrm{H}_2 \mathrm{~S}$.
$\therefore\left(\frac{15}{224}\right)$ moles of $\mathrm{O}_2$ uses $\left(\frac{2}{3} \times \frac{15}{22.4}\right)$ moles of $\mathrm{H}_2 \mathrm{~S}$
Molar mass of $\mathrm{H}_2 \mathrm{~S}=34 \mathrm{~g} / \mathrm{moI}$
$\therefore$ Gram of $\mathrm{H}_2 \mathrm{~S}=\left(34 \times \frac{2}{3} \times \frac{15}{224}\right) \mathrm{g}=15.17 \mathrm{~g}$
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