Nitric acid reacts with oxides

The added weight to $Y$ is the oxygen added to X hence $\displaystyle n_{O} = \frac{m_{Y}-m_{X}}{16} = \frac{2.71-2.23}{16} = 0.03 \text{ moles}$

$\displaystyle n_{NO_{2}} = \frac{0.672}{22.4} = 0.03 \text{ moles}$

Know that $2H^{+} + O^{2-} \to H_{2}O$

and $4H^{+} + NO_{3}^{-} \to NO + 2H_{2}O$

$\displaystyle \implies n_{HNO_{3}} = 2n_{O} + 4n_{NO_{2}} = 2 \times 0.03 + 4 \times 0.03 = \orange{\boxed{0.18}} \text{ moles}$