Question

Based on the following information: mass of Al is .91g, mass of Cu is 3.55g, and volume of 1 M CuSO₄ is 105 mL

What is the theoretical yield of Cu in moles, the theoretical yield of Cu in grams, the actual yield of Cu in grams, the percent yield of Cu, moles of CuSO₄, moles of Al, moles Cu product based on starting CuSO₄, and the moles Cu product based on starting Al.

Balanced Equation:

2AL(s) + 3CuSO₄(aq) à Al₂(SO₄)₃(aq) +3Cu(s)

Please show work

Answer 1

mass of Al = 0.91g

Moles of Al = mass/molecular weight

= 0.91g / 26.98g/mol

= 0.0337 mol

mass of Cu = 3.55 g = actual yield

Moles of Cu = 3.55g / 63.55g/mol

= 0.0533 mol

Moles of CuSO4 = molarity x volume

= 1 mol/L x 0.105 L = 0.105 mol

From the stoichiometry of the reaction

2 mol of Al required = 3 mol of CuSO4

0.0337 mol Al required = 3*0.0337/2

= 0.05055 mol of CuSO4

We have more moles of CuSO4 than required

Excess reactant = CuSO4

Limiting reactant = Al

Moles of Cu formed = 3 mol Cu x 0.0337 mol Al / 2 mol Al

= 0.05055 mol

Mass of Cu formed = moles x molecular weight

= 0.05055 mol x 63.55g/mol

= 3.212 g = theoretical yield

% yield = actual yield x 100 / theoretical yield

= 3.55 x 100 / 3.212

= 110.5%

There must be an error while calculating the mass of Cu in the experiment