Ch 202 Problem 3.76

Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and a solution of acetic acid. If 7.50 g sulfuric acid and 7.50 g lead (II) acetate are mixed, calculate the number of grams of sulfuric acid, lead (II) acetate, lead (II) sulfate, and acetic acid present in the mixture after the reaction is complete.

Yay!! This is a nice review of some earlier chapter 4 material, and is a good preview of coming attractions in in chapter 4! There's also some good chapter 2 review in here - you still need to know how to get from names to formulas!!!

Let's start with a balanced equation: you should be able to get this part yourself....

H2SO4(aq) + Pb(CH3COO)2(aq) ---> PbSO4(s) + 2CH3COOH(aq)

first, we need to find the limiting reactant. We convert g H2SO4 --> g lead (II) acetate and vice versa to see what we have.....

molar masses:

sulfuric acid

convert 7.50 g H2SO4 --> g lead (II) acetate:

need 24.9 g lead (II) acetate to consume 7.50 g sulfuric acid

convert 7.50 g lead (II) acetate --> g H2SO4

need 2.26 g H2SO4 to consume 7.50 g lead (II) acetate

OK, so lead (II) acetate is the limiting reactant - we need 24.9 g and we only have 7.50 g.

now that we know that lead (II) acetate is limiting, we can calculate the theoretical yields of both products by converting g lead (II) acetate to products.....

molar masses

acetic acid

theo. yield of lead (II) sulfate

theo.yield of acetic acid:

when the reaction is complete, there is NO lead (II) acetate remaining - it is limiting. We formed 6.99 g lead (II) sulfate, and 2.77 g acetic acid. We started with 7.50 g sulfuric acid, and we consumed (from above)

of the H2SO4. So,

so 5.24 g sulfuric acid remain, along with 6.99 g lead (II) sulfate and 2.77 g acetic acid.

DAC 7/3/08