1. Calculate moles of Zn: \frac{2.000 \text{ g}}{65.38 \text{ g/mol}} = 0.0306 \text{ mol} \text{ of Zn}
2. Calculate moles of CuSOβ: \frac{1.26 \text{ g}}{159.61 \text{ g/mol}} = 0.0079 \text{ mol} \text{ of CuSOβ}
3. Balanced equation: \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}
4. Limiting reactant is CuSOβ because it has fewer moles.
5. Moles of copper produced: 0.0079 mol (same as moles of CuSOβ).
6. Mass of copper produced: 0.0079 \text{ mol} \times 63.55 \text{ g/mol} = 0.50 \text{ g of Cu}