Question

What mass of precipitate can be made by reacting 5.00g of CaCl₂ (molar mass= 111 g/mol) with 5.00g of K₃PO₄ (molar mass = 212.3g/mol)?

Answer 1

The balanced equation is

2 K₃PO₄ + 3 CaCl₂ -------> Ca₃(PO₄)₂ + 6 KCl

Number of moles of CaCl2 = 5.00 g / 111 g/mol = 0.0450 mol

Number of moles of K3PO4 = 5.00 g / 212.3 g/mol = 0.0236 mol

From the balanced equation we can say that

2 mole of k3PO4 requires 3 mole of CaCl2 so

0.0236 mole of K3PO4 will require

= 0.0236 mole of K3PO4 *(3 mole of CaCl2 / 2 mole of K3PO4)

= 0.0354 mole of CaCl2

But we have 0.0450 mole of CaCl2 which is in excess so K3PO4 is limiting reactant

from the balanced equation we can say that

2 mole of K3PO4 produces 1 mole of Ca3(PO4)2 so

0.0236 mole of K3PO4 will produce

= 0.0236 mole of K3PO4 *(1 mole of Ca3(PO4)2 / 2 mole of K3PO4)

= 0.0118 mole of Ca3(PO4)2

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

the mass of 0.0118 mole of Ca3(PO4)2 = 3.66 g

Therefore, the mass of precipitate, Ca3(PO4)2 produced would be 3.66 g