Answer to Question #115391 in General Chemistry for ahmad

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) = 380.12 g/mol;

n=m/M;

n(Na₃PO₄·12H₂O) = (3.802 g)/(380.12 g/mol) = 0.01 moles;

Proportion according to the reaction:

2 moles (Na₃PO₄·12H₂O) — 1 mole (Ba₃(PO₄)₂)

0.01 moles (Na₃PO₄·12H₂O) — x moles (Ba₃(PO₄)₂)

x = 0.005 moles;

Answer: 0.005 moles