Question

For each of the following precipitation reactions, calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant.

Part A 2KI(aq)+Pb(NO3)2(aq)→PbI2(s)+2KNO3(aq) m = 17.3   g   SubmitMy AnswersGive Up Correct Part B Na2CO3(aq)+CuCl2(aq)→CuCO3(s)+2NaCl(aq) m =   g   SubmitMy AnswersGive Up Part C K2SO4(aq)+Sr(NO3)2(aq)→SrSO4(s)+2KNO3(aq) m =   g

Answer 1

Part a)

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

[(17.3 g Pb(NO₃)₂) / (331.2 g Pb(NO₃)₂/mol)] x [(2 mol KI /1 mol Pb(NO₃)₂) x (166 g KI/mol)] = 17.3 g KI

Part b)

Na₂CO₃(aq) + CuCl₂(aq)        CuCO₃(s) + 2NaCl(aq)

[(17.3 g CuCl₂)/(134.4 g CuCl₂/mol)] x [(1 mol Na₂CO₃/1 mol CuCl₂) x (105.9 g Na₂CO₃/mol)] = 13.6 g Na₂CO₃

Part c)

K₂SO₄(aq) + Sr(NO₃)₂(aq)          SrSO₄(s) + 2KNO₃(aq)

[(17.3 g Sr(NO₃)₂)/(211.6 g Sr(NO₃)₂/mol)] x [(1 mol K₂SO₄/1 mol Sr(NO₃)₂) x (174.2 g K₂SO₄/mol)] = 14.2 g K₂SO₄