Answer :
Given 5.00 g of Copper(II) chloride and 5.00 g of aluminum, 0.0372 moles of copper solid will be produced. Copper(II) chloride is the limiting reactant in this balanced chemical equation. The reaction produces moles of copper in a 1:1 ratio.
First, we need the balanced chemical equation for the reaction:
2Al + 3CuCl2 → 3Cu + 2AlCl3
Next, calculate the moles of each reactant:
- Molar mass of CuCl2 ≈ 134.45 g/mol
- Molecular mass of Al ≈ 26.98 g/mol
Moles of CuCl2: 5.00 g / 134.45 g/mol ≈ 0.0372 mol
Moles of Al: 5.00 g / 26.98 g/mol ≈ 0.185 mol
According to the stoichiometry of the reaction, 2 moles of Al react with 3 moles of CuCl2. Therefore, CuCl2 is the limiting reactant:
Moles of Cu produced: (0.0372 mol CuCl2) x (3 mol Cu / 3 mol CuCl2) ≈ 0.0372 mol
