What is the mass?

Question

A reaction between copper(I) chloride and lead(II) nitrate was run with 7.02 g of copper(I) chloride (molar mass = 99.00 g/mol) and 15.3 g of lead(II) nitrate (molar mass = 331.22 g/mol). If the percent yield of lead(II) chloride (molar mass = 278.1 g/mol) was 66.3%, how many grams of lead(II) chloride were actually produced?

Answer 1

To find mass of lead chloride (PbCl2), we must first find the theoretical yield.

2CuCl + Pb(NO₃)₂ ==> PbCl₂(s) + 2CuNO₃(aq) ... balanced equation

Is one of the reactants limiting? Let's check and see...

moles CuCl = 7.02 g CuCl x 1 mol CuCl/99.00 g = 0.0710 moles

moles Pb(NO₃)₂ = 15.3 g Pb(NO₃)₂ x 1 mol Pb(NO₃)₂ / 331.22 g = 0.0462 moles Pb(NO₃)₂

Because it takes 2 moles CuCl for each mol of Pb(NO₃)₂, the CuCl will be limiting.

Theoretical yield of PbCl₂ = 0.0710 mol CuCl x 1 mol PbCl₂/mol CuCl x 278.1 g/mol = 19.7 g PbCl₂

Actual yield with 66.3% yield = 19.7 g x 0.663 = 13.1 g PbCl₂