Question

When 3.802 g of Na3PO4·12H2O react with excess BaCl2·2H2O,
how many moles of Ba3(PO4)2 would be produced?

Answer 1

$2Na_3PO_4(aq)+3BaCl_2(aq) \rightarrow Ba_3(PO_4)_2(s)+6NaCl(aq)$

Convert grams of $Na_3PO_4 *12H_2O$ to moles

$3.802 g (\frac{1 mole}{380.18 g})=0.01000 mol$

Convert moles of $Na_3PO_4*12H_2O$ to moles of $Na_3PO_4$

$0.01000 mol Na_3PO_4*12H_2O (\frac{1 mole Na_3PO_4}{1 mole Na_3PO_4*12H_2O})=0.01000 mol Na_3PO_4$

Convert moles of $Na_3PO_4$ to moles of $Ba_3(PO_4)_2$

$0.01000 mol Na_3PO_4(\frac{1 mole Ba_3(PO_4)_2}{2 mole Na_3PO_4})=0.005 mol Ba_3(PO_4)_2$

Convert moles of $Ba_3(PO_4)_2$ to grams of $Ba_3(PO_4)_2$

$0.005 mol (\frac{601.92 g}{1 mole})=3.010 g Ba_3(PO_4)_2$