Calculate the theoretical yield of hydrogen gas (in l) at 295 k for a reaction of 1.50 g magnesium with excess hydrochloric acid. if 1.205 l of gas is produced, what is the percent yield of the reaction?

80.33 %.

Explanation:

Mg(s) + 2HCl(aq) → MgCl₂ + H₂.

n = mass/atomic mass = (1.50 g)/(24.3 g/mol) = 0.062 mol.

Using cross multiplication:

1.0 mol of Mg produces → 1.0 mol of hydrogen gas.

0.062 mol of Mg produces → 0.062 mol of hydrogen gas.

∵ PV = nRT.

∴ V of hydrogen (the theoretical yield) = nRT/P = (0.062 mol)(0.082 L.atm/mol.K)(295.0 K)/(1.0 atm) = 1.50 L.

The actual yield of the reaction = 1.205 L.

∴ The percent yield of the reaction = (actual yield)/(theoretical yield) x 100 = (1.205 L)/(1.50 L) x 100 = 80.33 %.