sulfuric acid. If 23.7 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)
Answers
Given info : 23.7 g of Ca₃(PO₄)₂ reacts with an excess of H₂SO₄. 13.4 of H₃PO₄ is produced.
To find : the percentage yield of the reaction is ..
solution : balanced chemical reaction of calcium phosphate and sulfuric acid is ..
Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 3H₃PO₄ (aq) + 3CaSO₄ (aq)
here you see, 3 moles of sulfuric acid reacts with one mole of calcium phosphate and it forms 3 moles of phosphoric acid.
∵ 3 moles of Ca₃(PO₄)₂ = 3 moles of H₃PO₄
⇒ 1 mole of Ca₃(PO₄)₂ = 1 mole of H₃PO₄
⇒ 310g of Ca₃(PO₄)₂ = 98 of H₃PO₄
⇒ 23.7 g of Ca₃(PO₄)₂ = 98/310 × 23.7 g of H₃PO₄ = 7.5 g of H₃PO₄
∵ percentage yield = theoretical value/ experimental value × 100
= 7.5g /13.4g × 100 = 56 %
therefore the percentage yield of the reaction is 56%.