mass of aluminum in the sample

Question

A mixture of aluminum and zinc, with a mass of 1.67 g, was dissolved in hydrochloric acid, and produced 1.69 liters of hydrogen gas, under standard conditions. Calculate the mass of aluminum in the sample.

Answer 1

2Al(s) + 6HCl(aq) ==> 2AlCl₃(aq) + 3H₂(g)

Zn(s) + 2HCl(aq) ==> ZnCl₂(aq) + H₂(g)

Under standard conditions, 1 mol H₂(g) = 22.4 L. Calculate moles of H₂(g) produced:

1.69 L H₂(g) x 1 mol H₂ / 22.4 L = 0.0754 mols H₂ produced in the overall reaction

Let X = g Al

Then 1.67 - X = g Zn

X / 26.98 = mols Al

1.67 - X / 65.4 = mols Zn

1 mol Al = 1.5 mol H₂

1 mol Zn = 1.0 mol H₂

Thus, 1.5 mol Al + 1 mol Zn = mol H₂

1.5•(X/26.98 g) + 1.0•(1.67-X/65.4 g) = 0.0754 mol H₂

X = 1.24 g Al