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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
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
5.68 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 :/
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:
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
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