A student heats 0.45 g of ethane, C₂H₆ (30 g/mol), with an excess of dry copper(II) oxide…

Step 1. $n(\text{C}_2\text{H}_6) = \dfrac{0.45\ \text{g}}{30\ \text{g/mol}} = 0.0150\ \text{mol}$.

Step 2. Carbonate. Two carbon atoms give two carbon dioxide molecules, so $n(\text{CO}_2) = 0.0300\ \text{mol}$. That gives $0.0300\ \text{mol}$ of $\text{CaCO}_{3}$ ($100\ \text{g/mol}$), which is $3.00\ \text{g}$.

Step 3. Water. Six hydrogen atoms give three water molecules, so $n(\text{H}_2\text{O}) = 0.0450\ \text{mol}$. The mass is $0.0450\ \text{mol} \times 18\ \text{g/mol} = 0.81\ \text{g}$.

Step 4. Copper(II) sulphate. Five water molecules for each $\text{CuSO}_{4}$: $\dfrac{0.0450}{5} = 0.00900\ \text{mol}$. The molar mass of $\text{CuSO}_{4}$ is $159.5\ \text{g/mol}$, so the mass is $0.00900 \times 159.5 = 1.44\ \text{g}$.