Finding Percent Yield

Question

Consider this reaction, which occurs in the atmosphere and contributes to photochemical smog:

P₂O₅(s) + 3H₂O(l) →2H₃PO₄(aq)

If there is 11.49 g P₂O₅ and excess H₂O present, the reaction yields 12.0 g H₃PO₄. Calculate the percent yield for the reaction.

Answer 1

Molar mass of P2O5 = 2×31+5×16 =142 g/mol

Molar mass of H3PO4 = 3+31+64 = 98 g/mol

11.49g P2O5 × 1mol P2O5/142g P2O5 × 2mol H3PO4 / 1mol P2O5 × 98 g H3PO4 / 1mol H3PO4 = 15.9 g H3PO4

(12.0 g H3PO4/15.9 g H3PO4) × 100% = 75.7 %

The percent yield for the reaction = 75.7%