Question

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)

Answer 1

1.       2KI(aq)     +     Pb(NO3)2(aq)   →    PbI2(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

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

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