If 23.6 g of Ca₃(PO₄)₂ reacts with an excess of H₂SO₄, what is the percent yield if 13.4 g of H₃PO₄ are formed in the following UNBALANCED chemical equation?

Question

If 23.6 g of Ca₃(PO₄)₂ reacts with an excess of H₂SO₄, what is the percent yield if 13.4 g of H₃PO₄ are formed in the following UNBALANCED chemical equation?

Ca₃(PO₄)₂ (s) + H₂SO₄ (aq) → H₃PO₄ (aq) + CaSO₄ (aq)

Answer 1

First, we MUST write a correctly balanced equation:

Ca₃(PO₄)₂ + 3H₂SO₄ ==> 2H₃PO₄ + 3CaSO₄ ... balanced equation

Use the stoichiometry of the balanced equation along with dimensional analysis to find the theoretical yield of H₃PO₄:

23.6 g Ca₃(PO₄)₂ x 1 mol Ca₃(PO₄)₂ / 310.2 g x 2 mol H₃PO₄ / 1 mol Ca₃(PO₄)₂ x 98.0 g H₃PO₄ / mol =

14.9 g H₃PO₄ theoretical yield

% yield = actual yield / theoretical yield (x100%) = 13.4 g / 14.9 g (x100%) = 89.9% yield