Answer to Question #221857 in General Chemistry for bbr

Question

1.50 g of aluminium reacted with 35.0 cm3 of 1.00 mol dm–3 copper(II) chloride solution, producing 1.98 g of solid copper. Calculate the percentage yield for this reaction.

2Al (s) + 3CuCl2 (aq) → 2AlCl3 (aq) + 3Cu (s)

Answer 1

Moles Al -Molar mass: 26.98g/mol-

1.50g Al * (1mol / 26.98g) = 0.0556 moles

Moles CuClâ‚‚:

35cm³ = 0.035L * (1.00mol / L) = 0.035 moles

For a complete reaction of Al there are necessaries:

0.0556 moles Al * (3 moles CuClâ‚‚ / 2 moles Al) = 0.0834 moles of CuClâ‚‚

As there are just 0.035 moles of CuClâ‚‚, CuClâ‚‚ is limiting reactant.

The solid produced is Cu, the theoretical mass of Cu is:

0.035 moles CuClâ‚‚ * (3 moles Cu / 3 moles CuClâ‚‚) = 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 =

89.2%