Question

Suppose the reaction ca3(PO4)2 + 3H2SO4 -> 3CaSO4 + 2H3PO4 is carried out starting with 103 g of Ca3(PO4)2 and 27.0 g of H2So4. How much phosphoric acid will be produced?

Answer 1

(103 g Ca3(PO4)2) / (310.1767 g Ca3(PO4)2/mol) = 0.33207 mol Ca3(PO4)2
(27.0 g H2SO4) / (98.0791 g H2SO4/mol) = 0.27529 mol H2SO4

0.27529 mole of H2SO4 would react completely with 0.27529 x (1/3) = 0.091763 mole of Ca3(PO4)2, but there is more Ca3(PO4)2 present than that, so Ca3(PO4)2 is in excess and H2SO4 is the limiting reactant.

(0.27529 mol H2SO4) x (2 mol H3PO4 / 3 mol H2SO4) x (97.9952 g H3PO4/mol) =
18.0 g H3PO4