Question in the description

Question

An impure sample of magnesium is treated with an excess of sulfuric acid, if 0.114 g of hydrogen is obtained from 4.750 g of the sample, calculate the percent purity (percent Mg) of the sample.

Answer 1

Mg(s) + H₂SO₄(aq) ==> MgSO₄(aq) + H₂(g) .. balanced equation

Find mols H₂ generated, and then from that, use the mol ratio of the balanced equation to find mols of Mg and then grams of Mg.

0.114 g H₂ x 1 mol H₂ / 2 g = 0.057 mols H₂

mols Mg = 0.057 mols H₂ x 1 mol Mg x 1 mol H₂ = 0.057 mols Mg

mass Mg = 0.057 mols Mg x 24.3 g / mol = 1.39 g Mg

% Mg = 1.39 g / 4.750 g (x100%) = 29.16% purity