When 13.7 g Ca(OH)2 reacts with 8.2 g H3PO4, how many grams of Ca3(PO4)2 are produced?

Question

3Ca(OH)2 + 2H3PO4 ---> Ca3(PO4)2 + 6H2O

Answer 1

3Ca(OH)₂ + 2H₃PO₄ ==> Ca₃(PO₄)₂ + 6H₂O ... balanced equation

Find limiting reactant. Easy way is to divide mols of each by corresponding coefficient

13.7 g Ca(OH)₂ x 1 mol Ca(OH)₂ / 70.1 g = 0.195 mols (÷3 -> 0.065)

8.2 g H₃PO₄ x 1 mol H₃PO₄ / 98.0 g = 0.0837 mols (÷2 -> 0.0419)

Because 0.0419 is less than 0.065, H₃PO₄ is limiting.

Grams Ca₃(PO₄)₂ produced:

0.0837 mol H₃PO₄ x 1 mol Ca₃(PO₄)₂ / 2 mol H₃PO₄ x 310 g Ca₃(PO₄)₂ / mol = 130. g Ca₃(PO₄)₂