Question

An aqueous solution containing 5.56 g of lead(II) nitrate is added to an aqueous solution containing 6.59g of potassium chloride.

The percent yield for the reaction is 84.5%. How many grams of precipitate is recovered? (in grams)

How many grams of the excess reactant remain? (in grams)

Answer 1

no of moles of Pb(NO3)2   = W/G.M.Wt

                                           = 5.56/331.2    = 0.01678moles

no of moles of KI = W/G.M.Wt

                              = 6.59/166   = 0.04moles

Pb(NO3)2(aq) + 2KI(aq) ---------------> PbI2(s) + 2KNO3(aq)

2 moles of KI react with 1 mole of Pb(NO3)2

0.04 moles of KI react with = 1*0.04/2 = 0.02 moles of Pb(NO3)2 is required

Pb(NO3)2 is limiting reactant

1 mole of Pb(NO3)2 react with excess of KI to gives 1 mole of PbI2

0.01678moles of Pb(NO3)2 react with excess of KI to gives 0.01678 moles of PbI2

mass of PbI2 = no of moles * gram molar mass

                    = 0.01678*461   = 7.74g

theoretical yield of PbI2 = 7.74g

percent yield    = actual yield *100/theoretical yield

     84.5           =   actual yield*100/7.74

actual yield    = 84.5*7.74/100   = 6.54g

precipitate is recovered    = 6.54g

Pb(NO3)2(aq) + 2KI(aq) ---------------> PbI2(s) + 2KNO3(aq)

1 mole of Pb(NO3)2 react with 2 mole of KI

0.01678 moles of Pb(NO3)2 react with = 2*0.01678/1 = 0.03356moles of KI

The no of moles of excess reactant remains after complete the reaction = 0.04-0.03356 = 0.00644moles

The amount of excess reactant remains after complete the reaction = no of moles * gram molar mass

                                                                                                            = 0.00644*166   = 1.07g >>>answer