Calculate the mass in grams of BaCl2 that will be required to produce 47.5 grams of Ba3(PO4)2.

Answer 1

What's the reaction taking place?

3BaCl₂(aq) + 2Na₃PO₄(aq) ==> Ba₃(PO₄)₂(s) + 6NaCl(aq) ... balanced equation

molar mass BaCl₂ = 208 g/mol

molar mass Ba₃(PO₄)₂ = 602 g/mol

47.5 g Ba₃(PO₄)₂ x 1mol / 602 g x 3 mol BaCl₂ / 1 mol Ba₃(PO₄)₂ x 208 g BaCl₂/mol = 49.2 g BaCl₂ needed