The reaction is as follows
$$
\mathrm{CaO}+\underset{126 \mathrm{~g}}{2 \mathrm{HNO}_3} \longrightarrow \underset{164 \mathrm{~g}}{\mathrm{Ca}\left(\mathrm{NO}_3\right)_2}+\mathrm{H}_2 \mathrm{O}
$$
Given, $56 \mathrm{~g} \quad 63 \mathrm{~g} \quad$ ?
According to balanced equation, 1 mole of $\mathrm{CaO}$ reacts with 2 moles of $\mathrm{HNO}_3$ to give 1 mole of $\mathrm{Ca}\left(\mathrm{NO}_3\right)_2$. But the given amount of $\mathrm{HNO}_3$ is exactly half, i.e. $1 / 2$ mole. In this reaction, $\mathrm{HNO}_3$ act as a limiting reagent.
As $126 \mathrm{~g} \mathrm{(2} \mathrm{moles)} \mathrm{of} \mathrm{HNO}_3 \longrightarrow 164 \mathrm{~g}$ ( $1 \mathrm{~mol}$ ) of $\mathrm{Ca}\left(\mathrm{NO}_3\right)_2$
So, $56 \mathrm{~g} \mathrm{(1} \mathrm{mol)} \mathrm{of} \mathrm{HNO}_3 \longrightarrow \frac{164}{2} \mathrm{~g}$ of $\mathrm{Ca}\left(\mathrm{NO}_3\right)_2$
$$
=82 \mathrm{~g} \text { of } \mathrm{Ca}\left(\mathrm{NO}_3\right)_2
$$