Question

$67.2$ litres of hydrogen combines with $44.8$ litres of nitrogen to form ammonia under specific conditions as:
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$
Calculate the volume of ammonia produced. What is the other substances, if any, that remains in the resultant mixture?

Answer 1

As seen in the reaction $1$ vol. of nitrogen combines with $3$ volumes of hydrogen to give $2$ volumes of ammonia.

It is done by the unitary method. As one mole of nitrogen is producing double the amount of ammonia, then the volume of ammonia is doubled as well. As per the mole concept, 1 mole is always equal to 22.4 L of any gas,

If $1$ vol. of $N_2$ gives $2$ vol. of $N{H}_3$.

Then, $44.8$ $l$ of $N_2$ gives = $2\times 44.8$ =$89.6$$l$.