##### 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

d) acetic acid

Jul 23rd, 2015

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

Jul 23rd, 2015

Here is my correct solution:

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

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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Jul 23rd, 2015
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Jul 23rd, 2015
Oct 19th, 2017
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