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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)76%89%37%56%

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)76%89%37%56%

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Solution

To calculate the percentage yield, we first need to determine the theoretical yield of copper.

  1. Calculate the number of moles of aluminium (Al) used: Aluminium has a molar mass of about 26.98 g/mol. So, the number of moles of Al = mass/molar mass = 1.50 g / 26.98 g/mol = 0.0556 mol.

  2. From the balanced chemical equation, we know that 2 moles of Al produce 3 moles of Cu. So, the number of moles of Cu produced = (0.0556 mol Al) * (3 mol Cu / 2 mol Al) = 0.0834 mol Cu.

  3. Convert this to grams using the molar mass of copper (Cu), which is about 63.55 g/mol. So, the theoretical yield of Cu = (0.0834 mol Cu) * (63.55 g/mol) = 5.30 g.

  4. The percentage yield = (actual yield / theoretical yield) * 100%. The actual yield is given as 1.98 g. So, the percentage yield = (1.98 g / 5.30 g) * 100% = 37.36%.

So, the closest answer is 37%.

This problem has been solved

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