How many grams of lead(ii) chloride is produced if 13.87 g lead(ii) nitrate combines with excess hydrochloric acid to produce lead(ii) chloride and nitric acid?

Pb(NO3)2+2HCl=PbCl2+2HNO3 Step 2 Find # of moles of Pb 13.87g Pb(NO3)2 x 1 mol/331.22=.041875 mol Pb(NO3)2 Step 3 Use balanced equation to find weight of Pb(NO3)2 .041875mol Pb x 1 mol Cl2/1mol Pb x 70.90g/1 mol Cl2=2.9689g Cl2 .041875 mol Pb x 207.2/1mol Pb=8.6765g Pb =11.65g

Particle exchange energy though elastic collision , hope this .