Question

The answer to each question of this set is a single digit integer, ranging from 0 to 9:

$HN{O}_3$ oxidises $N{H}_4^{+}$ ions to nitrogen and itself gets reduced to $N{O}_2$. The moles of $HN{O}_3$ required by 1 mol of $N{H}_4^{+}$ is

Answer 1

$(N{H}_4{)}_{2}C{O}_3+2HN{O}_3\to C{O}_2+H_{2}O+2N{H}_{4}N{O}_3$.

From the equatin we can conclude:

1 mole of $HN{O}_3$ is required by 1 mole of $N{H}_4^{+}$.