Amount needed to be mixed with excess amount of CaCl2 to produce 5.3 grams of percipitate

Question

Consider the chemical reaction :  2 Na₃PO₄ + 3 CaCl₂ → Ca₃(PO₄)₂ + 6 NaCl, what is the amount in grams of Na₃PO₄  needed to mixed with excess amount of CaCl₂ to produce 5.3 grams of the precipitate? 



The molar mass of Ca₃(PO₄)₂ : 310 g/mol

The molar mass of CaCl₂ : 111 g/mol

The molar mass of Na₃PO₄  : 164 g/mol

The molar mass of NaCl : 58 g/mol

Answer 1

5.3 g Ca₃(PO₄)₂ X (1 mol/310 g) X (2 mol Na3PO4/ 1mol Ca3PO4₂ ) X (164 g Na3Po4/1 mol )

= 5.608 g Na3Po4