A student treats 19.05 g of copper, with a molar mass of 63.5 g/mol, with dilute nitric acid until…

Step 1. Moles of copper. $\dfrac{19.05\ \text{g}}{63.5\ \text{g/mol}} = 0.300\ \text{mol}$.

Step 2. Nitric oxide. Three copper give two NO, so $0.300 \times \dfrac{2}{3} = 0.200\ \text{mol}$. The volume is $0.200\ \text{mol} \times 22.4\ \text{L/mol} = 4.48\ \text{L}$.

Step 3. Nitric acid used. Three copper use eight acid, so $0.300 \times \dfrac{8}{3} = 0.800\ \text{mol}$.

Step 4. Acid reduced. Each NO comes from one HNO3. So $0.200\ \text{mol}$ is reduced. The other $0.600\ \text{mol}$ becomes the nitrate in $\text{Cu(NO}_3)_2$: $0.300 \times 2 = 0.600\ \text{mol}$.