Starting with the following equation, BaCl₂(aq) + Na₃PO₄(aq) → Ba₃(PO₄)₂(s) + NaCl(aq) calculate the mass in grams of BaCl₂ that will be required to produce 41.5 grams of Ba₃(PO₄)₂.

Answer 1

First balance the reaction: 3 BaCl₂ + 2 Na₃PO₄ --> Ba₃(PO₄)₂ + 6 NaCl. Then [41.5 g Ba₃(PO₄)₂] * [1 mole Ba₃(PO₄)₂ / 601.93 g Ba₃(PO₄)₂] * [3 mole BaCl₂ / 1 mole Ba₃(PO₄)₂] * [208.23 g BaCl₂ / 1 mole BaCl₂] = 43.1 g BaCl₂.