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

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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
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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
ANSWER:
0g Pb(C2H3O2)25.68 g H2SO4
10.3 g PbSO4
4.06 g HC2H3O2
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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 :/
ok i will rework on the calculation
When will youn have the new answer?
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.
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Here is my correct solution:
SOLUTION:
H_{2}SO_{4}+ Pb(CH_{3}COO)_{2} = PbSO_{4} + 2CH_{3}COOH
mole H_{2}SO_{4 }= 9.00 g / 98.078 g/mol = 0.09176 moles
mole CH_{3}COOH = 9.00 g / 325.3 g/mol = 0.02767 moles
mole H_{2}SO_{4 }in excess = 0.09176  0.02767 = 0.06409 moles
mass H_{2}SO_{4 }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
ANSWER:
6.29 g H_{2}SO_{4}
0 g Pb(CH_{3}COO)_{2}
3.32 g CH_{3}COOH
8.39 g PbSO_{4}
P/S: Please remember to give me a 5 star review as said. Thanks again.
Thank you so much for your help, all of those answers were correct! :) You are a great tutor  I left U an amazing review
ok please left me a review. Thank you.
i still didn't see your review. please post it as said. Thanks.
Please keep you word as giving me a review for correct answers. Thanks.
I did, did you not see it?
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