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:

For the reaction:

Mg (s) + 2HCl (aq) → MgCl₂ + H₂.

Every 1.0 mole of Mg is dissolved in 2.0 moles of HCl and produce 1.0 mole of MgCl₂ and 1.0 mol H₂. To get the theoretical yield of hydrogen gas (in L) : We want to calculate the no. of moles of Mg in 1.50 g:

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 %.