Final answer:
Using the balanced equation and starting with 0.5 moles of copper sulphate as the limiting reactant, we calculate that 0.5 moles of copper will be produced, which corresponds to 31.775 grams of copper.
Step-by-step explanation:
The reaction between copper sulphate (CuSO4) and magnesium (Mg) to produce magnesium sulphate (MgSO4) and copper (Cu) is represented by the following balanced equation:
CuSO4 + Mg → MgSO4 + Cu
This stoichiometry tells us that one mole of copper sulphate reacts with one mole of magnesium to produce one mole of magnesium sulphate and one mole of copper. Since we are starting with 0.5 moles of copper sulphate and 1 mole of magnesium, copper sulphate is the limiting reactant, and will determine the amount of copper produced.
Therefore, 0.5 moles of copper sulphate will produce 0.5 moles of copper. To calculate the mass of copper produced, we use the molar mass of copper which is 63.55 g/mol:
0.5 moles Cu × 63.55 g/mol = 31.775 grams of Cu
The mass of copper produced from 0.5 moles of copper sulphate is 31.775 grams.