Q: Potassium chloride and lead (II) nitrite react according to the equation below. Determine the mass of lead (II) chloride produced if 168.9ml of 0.0250 M KCl is reacted with 250.0ml of 0.150 M lead (II) nitrite.

Flexi Says: Use the following steps to find the mass of product given the volume and molarity of a reactant:

1. Convert the amount of the reactant into moles using molarity (M) x volume (L).
2. Find the limiting reactant by calculating and comparing the amount of product each reactant will produce.
3. Use stoichiometry for the limiting reactant to find the mass of product produced.

Example: Determine the mass of lead (II) chloride produced if 168.9ml of 0.0250 M KCl is reacted with 250.0ml of 0.150 M lead (II) nitrite. The reaction is written as:
Pb(NO₂)₂ + 2 KCl = PbCl₂ + 2 KNO₂
From the reaction we find that 2 moles of KCl and 1 mol of Pb(NO₂)₂ react to produce 1 mole of PbCl₂. To find the number of moles in reaction we need to calculate moles by molarity x volume. For KCl = 4.2 moles , For Pb(NO₂)₂ = 37.5 moles. Here the limiting reagent is KCl, which produces one-half mole of PbCl₂. Therefore from 4.2 moles KCl the moles produced of PbCl₂ are 2.1 moles. The mass of of 2.1 mole PbCl₂ = 2.1 x 278.1 g/mol = 584.1 g.