Calculate the number of grams of each of the following present in the mixture:

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Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If 9.00 g of sulfuric acid and 9.00 g of lead(II) acetate are mixed, calculate the number of grams of each of the following present in the mixture after the reaction is complete:

a) Sulfuric Acid

b) lead(II) acetate

c) lead(II) sulfate

d) acetic acid

Jul 23rd, 2015

Thank you for the opportunity to help you with your question!

find the moles or H2SO4 and Pb(C2H3O2)2

9.00/98.086 = .091756 mol H2SO4

9.00/266.244 = .033804 mol Pb(C2H3O2)2

If you assume that the reaction proceeds to completion with equation

H2SO4 + Pb(C2H3O2)2 ===> PbSO4 + 2HC2H3O2

9 g H2SO4 = 9g / 98.08g/mole =0.09176 moles H2SO4
9 g HC2H3O2 = 9g / 266.24 g/mole = 0.033804 moles of Pb(OAc)2
After mixing-
0.02767 moles of PbSO4
0.05534 mole of HC2H3O2
0.00000 mole of Pb(C2H3O2)2
0.091756 - 0.033804 = 0.057952 moles of H2SO4
0.057952 * 98.08 g/mole = 5.68428 g H2SO4 

Then Pb(C2H3O2)2 must be the limiting reactant. There would remain 0 mol (and thus 0 g) of Pb(C2H3O2)2. There would remain .091756 - .033804 = .057952 mol H2SO4

convert to grams, so 0.057952*98.08 = 5.68428 g H2SO4

The ratio of Pb(C2H3O2)2 to PbSO4 is 1 mol to 1 mol, so there must be .033804 mol PbSO4 created (same as Pb(C2H3O2)2 used up). Convert to grams,

.033804 mol * (303.27g/mol) = 10.2517 g PbSO4

The ratio for HC2H3O2 is 1 mol Pb(C2H3O2)2 to 2 mol HC2H3O2 so there must be

.033804 mol Pb(C2H3O2)2 * [ 2 mol HC2H3O2 / 1 mol Pb(C2H3O2)2] notice the units cancel to leave

.067608 mol HC2H3O2

convert to grams:

.067608 mol * (60.052 g/mol) = 4.06000 g HC2H3O2


0g Pb(C2H3O2)2
5.68 g H2SO4
10.3 g PbSO4
4.06 g HC2H3O2

** P/S: Please remember to rate my answer later on once the answer is correct. I would be highly appreciate it. Thank you**
Jul 23rd, 2015

I already knew the H2SO4's correct answer, according to my teacher is: 

0.06409 moles x 98.08 g/mole = 6.29 g H2SO4 

So... I dont know what to say haha do you think you can rework evaluate it or something. Sorry :/

Jul 23rd, 2015

ok i will rework on the calculation

Jul 23rd, 2015

When will youn have the new answer?

Jul 23rd, 2015

I can send you the new answer very soon but can you please give a good review for this question? I would be appreciate that.

Jul 23rd, 2015

can you please give me a review now? if possible, i would be highly appreciate it and would definitely send the new correct answer very soon. Thanks for your help.

Jul 23rd, 2015

first give me answer :) and ill be happy to give u 5 stars

Jul 23rd, 2015

ok i will send the answer in 1-2 hours, but please keep your word. Thanks you.

Jul 23rd, 2015

Here is my correct solution:


H2SO4+ Pb(CH3COO)­2 = PbSO4 + 2CH3COOH

mole H2SO4 = 9.00 g / 98.078 g/mol = 0.09176 moles

mole CH3COOH = 9.00 g / 325.3 g/mol = 0.02767 moles

mole H2SO4 in excess = 0.09176 - 0.02767 = 0.06409 moles

mass H2SO4 in excess =0.06409 mol x 98.078 g/mol = 6.29 g

mass lead(II) acetate = 0

moles acetic acid = 2 x 0.02767 = 0.05534 moles

mass acetic acid = 0.05534 mol x 60.054 g/mol = 3.32 g

moles lead(II) sulfate = 0.02767 moles

mass lead(II) sulfate = 0.02767 mol x 303.26 g/mol = 8.39 g


6.29 g H2SO4

0 g Pb(CH3COO)­2

3.32 g CH3COOH

8.39 g PbSO4

Jul 23rd, 2015

P/S: Please remember to give me a 5 star review as said. Thanks again.

Jul 23rd, 2015

Thank you so much for your help, all of those answers were correct! :) You are a great tutor - I left U an amazing review

Jul 24th, 2015

ok please left me a review. Thank you.

Jul 24th, 2015

i still didn't see your review. please post it as said. Thanks.

Jul 24th, 2015

Please keep you word as giving me a review for correct answers. Thanks.

Jul 24th, 2015

I did, did you not see it?

Jul 24th, 2015

no i didn't see any review from you.

Jul 24th, 2015

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