Question

Attempt 2 - An aqueous solution containing 8.03 g of lead(II) nitrate is added to an aqueous solution containing 6.33 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. balanced chemical equation: Pb(NO3)2(aq) + 2Cl(aq) PbCI,(s) + 2KNO, (aq) What is the limiting reactant? O potassium chloride O lead(II) nitrate The percent yield for the reaction is 89.3%. How many grams of precipitate is recovered? precipitate recovered: How many grams of the excess reactant remain? excess reactant remaining: 2.72

Answer 1

Giuan balanced chemicalreactien i3 Pbus +2 KNos Cav s) +2 Kd Can) t (Nos) (ay rdar Masty ef ptNos 331 2 9mmol 74 55 9mlmol Kd 1 278.1gm/mbl Pbuy 1mol of pNo, reaulsith 2moles ef K IX 331 29mo PbNog),reautsith 2x7455 Fnet 8.039m of Ked. 2 1 1 11 8.03 2x 74. 55 = 3-619ne Ky 331-2 given 6-33g ef kul, So Excess reagent = Ku miting one = pb(Nos Percent yield- 89.3 Imol ef pb(Nos 3312 9m of phNo3)2 8.03 9om 1mol ef Pbus 278.1m of Pbu, 2 6-74 Pmef Pbl2 formed 8.03 X 2 78.1 331 2 Armount pre cipitak recavered 674.x89.3 100 -6-029m8 Excess reactant (ku) remain = 6.33-3-61 = 2-72 9m3