An industrial vat contains 650 grams of solid lead(ii) chloride formed from a reaction of 870 grams of lead(ii) nitrate with excess hydrochloric acid. this is the equation of the reaction: 2hcl + pb(no3)2 → 2hno3 + pbcl2. what is the percent yield of lead(ii) chloride? the percent yield of lead chloride is %

             88.98 %

Solution:

The Balance Chemical Equation is as follow,

                                   2 HCl + Pb(NO₃)₂     →    2 HNO₃ + PbCl₂

According to equation,

           331.2 g (1 mole) Pb(NO₃)₂ produces  =  278.1 g (1 mole) PbCl₂

So,

                  870 g of Pb(NO₃)₂ will produce  =  X g of PbCl₂

Solving for X,

                      X  =  (870 g × 278.1 g) ÷ 331.2 g

                      X  =  730.5 g of PbCl₂

Therefore,

                Theoretical Yield =  730.5 g

Also as given,

                 Actual Yield =  650 g

So using following formula for percentage yield,

                         %age Yield  =  (Actual Yield / Theoretical Yield) × 100

Putting values,

                         %age Yield  =  (650 g / 730.5 g) × 100

                         %age Yield  =  88.98 %

not clear