For each of the following precipitation reactions, calculate how many grams of the first reactant are necessary to completely react with 18.0 g of the second reactant.
a. 2KI(aq)+Pb(NO3)2(aq)→PbI2(s)+2KNO3(aq)
b. Na2CO3(aq)+CuCl2(aq)→CuCO3(s)+2NaCl(aq)
c. K2SO4(aq)+Sr(NO3)2(aq)→SrSO4(s)+2KNO3(aq)
Mass: 2 x 166 g 331.2 g 461 g 2 x 101.1 g
331.2 g of Pb(NO3)2 requires KI = 2 x 166 g
18 g of Pb(NO3)2 requires KI = ( 2 x 166 x 18)/331.2 g = 18.04 g
Mass: 106 g 134.5 g 123.6 g 2 x 58.4 g
134.5 g of CuCl2 requires Na2CO3 = 106 g
18 g of CuCl2 requires Na2CO3 = (106 x 18)/134.5 g = 14.2 g
(c) K2SO4(aq) + Sr(NO3)2(aq) → SrSO4(s) + 2KNO3(aq)
Mass: 174.3 g 211.6 g 183.7 g 2 x 101.1 g
211.6 g of Sr(NO3)2 requires K2SO4 = 174.3 g
18 g of Sr(NO3)2 requires K2SO4 = (174.3 x 18)/211.6 g = 14.8 g
For each of the following precipitation reactions, calculate how many grams of the first reactant are...
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