1. When 0.990g of Na3PO4.12H20 reacts with excess BaCl2.2H20 how many
moles of Ba3(PO4)2are produced?​

The balanced reaction is

(Na3PO4)12H2O + 9(BaCl2)2H2O → 6Ba3(PO4)2 + 36NaCl + 10H2O

molar mass of (Na3PO4)12H2O = 380 g/mol

given mass of (Na3PO4)12H2O = 0.990 g

moles of (Na3PO4)12H2O = given mass/molar mass

moles of (Na3PO4)12H2O = 0.990/380= 0.0026 mole

one mole of (Na3PO4)12H2O forms 6 mole of Ba3(PO4)2

0.0026 mole of (Na3PO4)12H2O forms 0.0026×6 =

0.0156 mole of Ba3(PO4)2

mass of 0.0156 moles of Ba3(PO4)2 is = 0.0156 × 601.9 g = 9.40 gram

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explanation:

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