Question

Write a balanced metathesis equation for the reaction which takes place when the two solutions described below are combined and then determine the amount of solid product produced.

300.0 ml of a 0.200 M Na3PO4 solution is combined with 300.0 ml of a 0.500M Pb(NO3)2 solution.How many grams of solid lead phosphate will be produced by the reaction. Assume 100% yield.

Answer 1

Ans. Balanced reaction: 3 Pb(NO₃)₂(aq) + 2 Na₃PO₄(aq) ---> Pb₃(PO₄)₂(s) + 6 NaNO₃(aq)

Theoretical molar ratio of reactants = Pb(NO₃)₂ : Na₃PO₄ = 3 : 2 = 1.5 : 1

#. # Moles of Na₃PO₄ = Molarity x Volume of solution in liters

                                    = 0.200 M x 0.300 L

                                    = 0.06 mol

Moles of Pb(NO₃)₂ = 0.500 M x 0.300 L = 0.15 mol

Experimental molar ratio of reactants = Pb(NO₃)₂ : Na₃PO₄ = 0.15 : 0.06 = 2.5 : 1

Compare the theoretical and experimental molar ration of reactants, the moles of Pb(NO₃)₂ is greater than 1.5 when that of NaNO₃ is kept constant at 1 mol.

Therefore,

                        Pb(NO₃)₂ is the reagent in excess

Na₃PO₄ is the limiting reactant.

The formation of product follows the stoichiometry of limiting reactant.

#. According to stoichiometry of balanced reaction, 2 mol NaNO₃ (limiting reactant) produces 1 mol Pb₃(PO₄)₂. So,

            Moles of Pb₃(PO₄)₂ produced = (1/2) x Moles of NaNO₃

                                                            = (1/2) x 0.06 mol

                                                            = 0.03 mol

Mass of Pb₃(PO₄)₂ produced = Moles x Molar mass

                                                            = 0.03 mol x (811.542724 g/ mol)

                                                            = 24.35 g