Question

When sodium bicarbonate is heated, $1.806\times {10}^{24}$ molecules of water is obtained. Find the volume of $C{O}_2\left(g\right)$ obtained at STP.
$2NaHC{O}_3\stackrel{\Delta }{\to }N{a}_{2}C{O}_3+C{O}_2+H_{2}O$

• A. $67.2\text{L}$
• B. $55.0\text{L}$
• C. $60.7\text{L}$
• D. $70.5\text{L}$

Answer 1

The correct option is A $67.2\text{L}$

$2NaHC{O}_3\stackrel{\Delta }{\to }N{a}_{2}C{O}_3+C{O}_2+H_{2}O$

Moles of $H_{2}O=\frac{1.806\times {10}^{24}}{6.02\times {10}^{23}}=3\text{mol}$

2 mol of $NaHC{O}_3$ decompose to form 1 mol of $C{O}_2$ and 1 mol of $H_{2}O$.
As per the reaction, if 3 mol of $H_{2}O$ is getting formed, moles of $C{O}_2$ formed will also be 3 mol.
Hence, volume of $C{O}_2$ obtained at STP $=3\times 22.4\text{L}=67.2\text{L}$