Answer is: The percent yield of lead chloride is 88%.
Balanced chemical reaction: 2HCl + Pb(NO₃)₂ → 2HNO₃ + PbCl₂.
m(PbCl₂) = 650 g; mass of lead (II) chloride.
m(Pb(NO₃)₂) = 870 g; mass of lead (II) nitrate.
n(Pb(NO₃)₂) = 870 g ÷ 331.21 g/mol.
n(Pb(NO₃)₂) = 2.63 mol; amount of lead (II) nitrate.
From balanced chemical reaction: n(Pb(NO₃)₂) : n(PbCl₂) = 1 : 1.
n(PbCl₂) = 2.63 mol; amount of lead (II) chloride.
m(PbCl₂) = 2.63 mol · 278.1 g/mol.
m(PbCl₂) = 731.4 g.
yield = 650 g ÷ 731.4 g · 100%.
yield = 88.88 %.