Solve stoichiometry problem

Question

A 50.0g sample of calcium carbonate is reacted with 35.0g phosphoric acid to produce calcium phosphate and carbonic acid (H2CO3(aq)). Determine:

1. The number of grams of calcium phosphate that could be produced.
2. The number of grams of excess reagent that will remain.

Answer 1

Balance the reaction to get mole ratios: 3 CaCO₃ + 2 H₃PO₄ --> Ca₃(PO₄)₂ + 3 H₂CO₃. Compare yields: (50.0g CaCO₃)*(1 mol CaCO₃ /100.1g CaCO₃)*(1 mol Ca₃(PO₄)₂ /3 mol CaCO₃) = 0.166₅ mole Ca₃(PO₄)₂.

(35.0g H₃PO₄)*(1 mol H₃PO₄ /97.99g H₃PO₄)*(1 mol Ca₃(PO₄)₂ /2 mol H₃PO₄) = 0.178₆ mole Ca₃(PO₄)₂.

1) CaCO₃ is limiting, producing (0.166₅ mol Ca₃(PO₄)₂)*(310.2g /1 mol Ca₃(PO₄)₂) = 51.7g Ca₃(PO₄)₂.

2) [Δmole = 0.012₁ mol Ca₃(PO₄)₂]*[2 mol H₃PO₄ /1 mol Ca₃(PO₄)₂]*(97.99g H₃PO₄ /1 mol H₃PO₄) = 2.4g H₃PO₄.