Question

When 3.802 g of Na3 PO4 ·12H2O react with excess BaCl2 ·2H2O, how many moles of Ba3 (PO4 ) 2 would be produced?

Answer 1

2Na₃PO₄·12H₂O + 3BaCl₂·2H₂O → Ba₃(PO₄)₂ + 6NaCl + 30H₂O

M(Na₃PO₄·12H₂O) = 380.124 g/mol

n(Na₃PO₄·12H₂O) $= \frac{3.802}{380.124}$ = 0.01 mol

n(Ba₃(PO₄)₂) $= \frac{1}{2}$ n(Na₃PO₄·12H₂O) $= \frac{1}{2} \times 0.01$ = 0.005 mol

Answer: 0.005 mol