Question

If 1.50 mol of CaCl2 is mixed with 1.5 mol Na3PO4, the maximum amount in moles of Ca3(PO4)2 that can be formed is 2 Na3PO4 + 3CACl2 -> Ca3(PO4)2 + 6 NaCl

Answer 1

From the balanced equation we can say that

2 mole of Na3PO4 requires 3 mole of CaCl2 so

1.50 mole of Na3PO4 will require

= 1.50 mole of Na3PO4 *( 3 mole of CaCl2 / 2 mole of Na3PO4)

= 2.25 mole of CaCl2

But we have only 1.50 mole of CaCl2 which is in short so CaCl2 is limiting reactant

From the balanced equation we can say that

3 mole of CaCl2 produces 1 mole of Ca3(PO4)2 so

1.50 mole of CaCl2 will produce

= 1.50 mole of CaCl2 *(1 mole of Ca3(PO4)2 / 3 mole of CaCl2)

= 0.500 mole of Ca3(PO4)2

mass of 1 mole of Ca3(PO4)2 = 310.18 g so

the mass of 0.500 mole of Ca3(PO4)2 = 155 g

Therefore, the mass of Ca3(PO4)2 produced would be 155 g