Percentage Yield

Question

If 26.6 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 + H₂SO₄ → H₃PO₄ + CaSO₄

Answer 1

You need a balanced equation.

Ca₃(PO₄)₂ + H₂SO₄ → H₃PO₄ + CaSO₄

You have 3 Ca on the left but only 1 on the right. So you need to add 3 to the right. But now you have 3 SO₄ so you need add 3 to the left .

Ca₃(PO₄)₂ + 3 H₂SO₄ → H₃PO₄ + 3 CaSO₄

Now you have 6 Hs on the left, but only 3 Hs on the right. So you need to add 2 to the right. Now every thing is balanced. Double check to make sure you understand this.

Ca₃(PO₄)₂ + 3 H₂SO₄ → 2 H₃PO₄ + 3 CaSO₄

This equation says 1 mol of Ca₃(PO₄)₂ reacts with 3 mols of H₂SO₄ to form 2 mols of H₃PO₄ and 3 mols of CaSO₄

The reactant amount given is 26.6 g of Ca₃(PO₄)₂ which needs to be turned into mols. If all of this mol amount of Ca₃(PO₄)₂ is converted to the product, then that gives you the theoretical yield.

The product amount given is 13.4 g of H₃PO₄ which is the actual yield in grams. The percent yield is (actual yield) ÷ (theoretical yield)

You will now convert the grams of Ca₃(PO₄)₂ molecular mass which is 310.18 g/mol

But you know 1 mol of Ca₃(PO₄)₂ reacts to form 2 mols of H₃PO₄ which has molecular mass of 97.99 g/mol

Convert the gram of Ca₃(PO₄)₂ to mols

26.6 g ( 1mol / 310.18 g) = 0.08575 mol of Ca₃(PO₄)₂

Use the molar ration from the balanced equation.

0.08575 mol of Ca₃(PO₄)₂ (2mol H₃PO₄ / 1 mol Ca₃(PO₄)₂ ) = 0.1715 mol of H₃PO₄

But H₃PO₄ has mol mass of 97.99 g/mol which means, you can theoreitcally make

0.1715 mol of H₃PO₄ (97.99g/1 mol H₃PO₄ ) = 16.80g of H₃PO₄

The actual yield was 13.4 g which means the percent yield was

13.4g/16.8g = 0.797 or 79.7% was the yield.