JEE MAIN - Chemistry (2022 - 25th July Evening Shift - No. 14)
56.0 L of nitrogen gas is mixed with excess hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _________ L.
Answer
46
Explanation
$$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})$$
Since $$\mathrm{H}_{2}$$ is in excess and $$20 \mathrm{~L}$$ of ammonia gas is produced.
Hence, 2 moles $$\mathrm{NH}_{3} \equiv 1$$ mole $$\mathrm{N}_{2} \quad(v \propto \mathrm{n})$$
$$ 20 \mathrm{~L} \mathrm{NH}_{3} \equiv 10 \mathrm{~L} \mathrm{~N}_{2} $$
Volume of $$\mathrm{N}_{2}$$ left $$=56-10$$
$$ =46 \mathrm{~L} $$
Since $$\mathrm{H}_{2}$$ is in excess and $$20 \mathrm{~L}$$ of ammonia gas is produced.
Hence, 2 moles $$\mathrm{NH}_{3} \equiv 1$$ mole $$\mathrm{N}_{2} \quad(v \propto \mathrm{n})$$
$$ 20 \mathrm{~L} \mathrm{NH}_{3} \equiv 10 \mathrm{~L} \mathrm{~N}_{2} $$
Volume of $$\mathrm{N}_{2}$$ left $$=56-10$$
$$ =46 \mathrm{~L} $$
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