Answer:
89.2%
Explanation:
The reaction of Al with CuCl₂ is:
2Al(s) + 3CuCl₂(aq) → 2AlCl₃(aq) + 3Cu(aq)
Assuming the reaction occurs until limiting reactant is over, we must find limiting reactant:
Moles Al -Molar mass: 26.98g/mol-
1.50g Al * (1mol / 26.98g) = 0.0556 moles
Moles CuCl₂:
35cm³ = 0.035L * (1.00mol / L) = 0.035 moles
For a complete reaction of Al there are necessaries:
0.0556 moles Al * (3 moles CuCl₂ / 2 moles Al) = 0.0834 moles of CuCl₂
As there are just 0.035 moles of CuCl₂, CuCl₂ is limiting reactant.
The solid produced is Cu, the theoretical mass of Cu is:
0.035 moles CuCl₂ * (3 moles Cu / 3 moles CuCl₂) = 0.035 moles Cu
0.035 moles Cu * (63.546g / mol) = 2.22g Cu
As actual yield was 1.98g, percentage yield is:
1.98g Cu / 2.22g Cu * 100 =
