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

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 =

hi my dude or dudette