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 (integer)
46
Solution
$$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})$$
<br/><br/>
Since $\mathrm{H}_{2}$ is in excess and $20 \mathrm{~L}$ of ammonia gas is produced.
<br/><br/>
Hence, 2 moles $\mathrm{NH}_{3} \equiv 1$ mole $\mathrm{N}_{2} \quad(v \propto \mathrm{n})$
<br/><br/>
$20 \mathrm{~L} \mathrm{NH}_{3} \equiv 10 \mathrm{~L} \mathrm{~N}_{2}$
<br/><br/>
Volume of $\mathrm{N}_{2}$ left $=56-10$
<br/><br/>
$=46 \mathrm{~L}$
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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