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$2 \mathrm{KHCO}_3 \longrightarrow \ldots+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$ find amount of gases formed (in lit).
When amount of $\mathrm{KHCO}_3$ is 33 gm .
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When amount of $\mathrm{KHCO}_3$ is 33 gm .
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Verified Answer
The correct answer is:
7.46
The given reaction is as follows
$2 \mathrm{KHCO}_3 \longrightarrow \mathrm{K}_2 \mathrm{CO}_3(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}$ (l)
$\because 2$ moles of $\mathrm{KHCO}_3$ gives $=1 \mathrm{~mol}$ of $\mathrm{K}_2 \mathrm{CO}_3$
$\therefore 1$ moles of $\mathrm{KHCO}_3$ gives $=1 / 2 \mathrm{~mol}$ of $\mathrm{K}_2 \mathrm{CO}_3$
$\therefore 33 \mathrm{gm}\left(\frac{33}{100} \mathrm{~mol}\right)$ of $\mathrm{KHCO}_3$ gives $=\frac{33}{200}$
mol of $\mathrm{K}_2 \mathrm{CO}_3$
Similarly,
$33 \mathrm{gm}\left(\frac{33}{100} \mathrm{~mol}\right)$ of $\mathrm{KHCO}_3$ gives $=\frac{33}{200} \mathrm{~mol}$ of $\mathrm{CO}_2$
Total moles of gases $=\frac{33}{200}+\frac{33}{200}=\frac{66}{200} \mathrm{~mol}$
$\begin{aligned} 1 \mathrm{~mol} \text { of gas } & =224 \mathrm{l} \text { volume } \\ \therefore \quad \frac{66}{200} \mathrm{~mol} \text { of gas } & =\frac{66}{200} \times 22.4=7.46 \mathrm{l} .\end{aligned}$
$2 \mathrm{KHCO}_3 \longrightarrow \mathrm{K}_2 \mathrm{CO}_3(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}$ (l)
$\because 2$ moles of $\mathrm{KHCO}_3$ gives $=1 \mathrm{~mol}$ of $\mathrm{K}_2 \mathrm{CO}_3$
$\therefore 1$ moles of $\mathrm{KHCO}_3$ gives $=1 / 2 \mathrm{~mol}$ of $\mathrm{K}_2 \mathrm{CO}_3$
$\therefore 33 \mathrm{gm}\left(\frac{33}{100} \mathrm{~mol}\right)$ of $\mathrm{KHCO}_3$ gives $=\frac{33}{200}$
mol of $\mathrm{K}_2 \mathrm{CO}_3$
Similarly,
$33 \mathrm{gm}\left(\frac{33}{100} \mathrm{~mol}\right)$ of $\mathrm{KHCO}_3$ gives $=\frac{33}{200} \mathrm{~mol}$ of $\mathrm{CO}_2$
Total moles of gases $=\frac{33}{200}+\frac{33}{200}=\frac{66}{200} \mathrm{~mol}$
$\begin{aligned} 1 \mathrm{~mol} \text { of gas } & =224 \mathrm{l} \text { volume } \\ \therefore \quad \frac{66}{200} \mathrm{~mol} \text { of gas } & =\frac{66}{200} \times 22.4=7.46 \mathrm{l} .\end{aligned}$
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