Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid.5.80 g of sulfuric acid and 5.80 g of lead(II) acetate are mixed.

Part A

Calculate the number of grams of sulfuric acid present in the mixture after the reaction is complete.

m =

  g  

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Part B

Calculate the number of grams of lead(II) acetate present in the mixture after the reaction is complete.

m =

  g  

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Part C

Calculate the number of grams of lead(II) sulfate present in the mixture after the reaction is complete.

m =

  g  

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Part D

Calculate the number of grams of acetic acid present in the mixture after the reaction is complete.

Answer 1

H2SO4(aq) + PbAc2(aq) = PbSO4(s) + 2HAc(aq)

m = 5.8 g of H2SO4

mol = mass/MW = 5.8/98 = 0.05918367346 mol acid

m = 5.8 g of PbAc

PbAc = masS/MW = 5.8/325.29 = 0.017830 mol of PbAc

the ratio is

1:1

for H2SO4:

final mol = initial - reacted = 0.05918367346- 0.017830 = 0.04135367346mol of H2SO4

mass = mol*MW = 0.04135367346*98 = 4.052 g of H2SO4 left

b)

there are no grams of Lead Acetetate left (all react)

c)

sinceratio is 1:1

then

0.017830 mol of PbSO4 aer formed

or

mass = mol*MW = 0.017830 *303.26 = 5.4071 g of PbSO4

d)

for Acetic acid

ratio is

1:2

so0.017830 *" = 0.03566 mol of HAc

WM of HAc = 60

then

mass = mol*MW = 0.0356660 = 2.1396g of HAc