A piece of copper of mass 12.7 g (63.5 g/mol) is warmed with an excess of concentrated sulphuric…

Step 1. Copper. $\dfrac{12.7\ \text{g}}{63.5\ \text{g/mol}} = 0.200\ \text{mol}$.

Step 2. Sulphur dioxide. One mole of Cu gives one mole of SO2: $0.200 \times 22.4 = 4.48\ \text{L}$.

Step 3. The acid. Two moles of acid for each mole of copper: $0.400\ \text{mol}$.

Step 4. Copper sulphate. $0.200 \times 159.5 = 31.9\ \text{g}$.

Step 5. The oxidising agent. Sulphur goes from $+6$ in $\text{H}_2\text{SO}_4$ to $+4$ in $\text{SO}_2$ for $0.200\ \text{mol}$ of the acid. The other $0.200\ \text{mol}$ becomes the sulphate of $\text{CuSO}_4$, with sulphur still at $+6$.