Question

Determine the mass (in g) of Ca₃(PO₄)₂ that is produced when 140 mL of a 8.48×10^-2 M Na₃PO₄ solution completely reacts with 778 mL of a 1.54×10^-2 M CaCl₂ solution according to the following balanced chemical equation.  

2Na₃PO₄(aq) + 3CaCl₂(aq) → Ca₃(PO₄)₂(s) + 6NaCl(aq)

Answer 1

Number of moles of Na₃PO₄ , n = molarity x volume in L

                                              = 8.48x10-2 M x 0.140 L

                                              = 0.0119 moles

Number of moles of CaCl₂ , n' = Molarity x volume in L

                                            = 1.54x10-2 M x 0.778 L

                                            = 0.0120 moles

Molar mass of Ca₃(PO₄)₂ =(3x40)+(2x31)+(8x16) = 310 g/mol

2Na₃PO₄(aq) + 3CaCl₂(aq) → Ca₃(PO₄)₂(s) + 6NaCl(aq)

According to the balanced equation ,

2 moles of Na₃PO₄ reacts with 3 moles of CaCl₂

N moles of Na₃PO₄ reacts with 0.0120 moles of CaCl₂

N = (2x0.0120) / 3

= 0.008 moles

So 0.0119 - 0.008 = 0.0039 moles of Na₃PO₄ left unreacted so it is the excess reactant.

Since all the mass of CaCl₂ completly reacted it is the limiting reactant.

According to the balanced equation,

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

0.0120 moles of CaCl₂ produces M mole of Ca₃(PO₄)₂

M = (1x0.0120) / 3

   = 0.004 moles x310 (g/mol)

   = 1.24 g

Therefore the mass of Ca₃(PO₄)₂ produced is 1.24 g

So mass of Ca₃(PO₄)₂ is , m = number of moles x molar mass

                                          = 0.004 mol x