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Question

30.00 mL of 0.200 M Pb(NO₃)₂(aq) are mixed with 25.00 mL of 0.100 M K₃PO₄(aq). 1. What is the formula of the solid product formed in this reaction? 2. What mass of this solid product will be formed?

Answer 1

3Pb(NO₃)₂(aq) + 2K₃PO₄(aq) ==> Pb₃(PO₄)₂(s) + 6KNO₃(aq) ... balanced equation

1. Formula of solid product is Pb₃(PO₄)₂(s)
2. Find the limiting reactant:

0.030 L x 0.200 mol/L = 0.006 moles Pb(NO₃)₂ (÷3-->0.002)

0.025 L x 0.100 mol/L = 0.0025 moles K₃PO₄ (÷2-->0.00125)

Limiting reactant is K₃PO₄

mass Pb₃(PO₄)₂(s) = 0.0025 mol K₃PO₄ x 1 mol Pb₃(PO₄)₂/2 mol K₃PO₄ x 812 g/mol = 1.02 g (3 sig figs)