Q11 / 160 · AP EAMCET 2021 (23 Aug Shift 1) · Chemistry
For formation of $3.40 \mathrm{~g}$ of ammonia gas, what volumes of hydrogen gas and nitrogen gas, respectively are required at NTP conditions?
- A$2.24 \mathrm{~L}$ and $2.24 \mathrm{~L}$
- B$2.24 \mathrm{~L}$ and $1.24 \mathrm{~L}$
- Correct answerC$6.72 \mathrm{~L}$ and $2.24 \mathrm{~L}$
- D$6.72 \mathrm{~L}$ and $1.12 \mathrm{~L}$
Solution
$$
\begin{array}{ccc}
\text { (c) } \mathrm{N}_2 & +3 \mathrm{H}_2 & \underset{1 \mathrm{~V}}{\longrightarrow} \mathrm{NH}_3 \\
22.4 \mathrm{~L} & 22.4 \times 3 & 2 \mathrm{~V} \\
=224 \mathrm{~L} & =67.2 \mathrm{~L} & =44.8 \mathrm{~L}
\end{array}
$$
According to Gay-Lussac's law, 3 volumes of $\mathrm{H}_2$ will give 2 vol. of $\mathrm{NH}_3$.
$$
\therefore 67.2 \mathrm{~L} \mathrm{of} \mathrm{H}_2 \text { will give }=\frac{2 \times 67.2}{3}=44.8 \mathrm{~L} \mathrm{of} \mathrm{NH}_3
$$
At the same time, 3 vol. of $\mathrm{H}_2$ will react $=1$ vol. of $\mathrm{N}_2$.
$$
\begin{aligned}
\therefore 67.2 \mathrm{~L} \mathrm{of} \mathrm{H}_2 \text { will react } & =\frac{1}{3} \times 67.2 \mathrm{~L} \mathrm{of} \mathrm{N}_2 \\
& =224 \mathrm{~L} \text { of } \mathrm{N}_2
\end{aligned}
$$
67.2 $\mathrm{L}$ of $\mathrm{H}_2$ and $22.4 \mathrm{~L}$ of $\mathrm{N}_2$ are required to form $34 \mathrm{~g}$ of $\mathrm{NH}_3$.
So, for $3.4 \mathrm{~g}$ of $\mathrm{NH}_3$
6.72 $\mathrm{L}$ and $2.24 \mathrm{~L}$ of $\mathrm{H}_2$ and $\mathrm{N}_2$ gas are required respectively.
Hence, the correct answer is option: 'C'.