Question

1.

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

Answer 1

0.990g of Ba(NO3)2 will produce approximately 0.0047477 moles of Ba3(PO4)2

Step-by-step explanation:

The balanced chemical equation for the reaction is:

3Ba(NO3)2(aq) + 2Na3PO4(aq) → Ba3(PO4)2(s) + 6NaNO3(aq)

From the balanced equation, we can see that for every 3 moles of Ba(NO3)2 that react, 1 mole of Ba3(PO4)2 is produced. Therefore, the number of moles of Ba3(PO4)2 produced will be equal to:

(0.990g Ba(NO3)2) / (261.336g/mol Ba(NO3)2) * (1 mol Ba3(PO4)2) / (3 mol Ba(NO3)2) ≈ 0.0047477 moles

Answer 2

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