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 22nd, 2015

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

Equation:

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

Now: 
9 g H2SO4 = 9g / 98.08g/mole 

=0.09176 moles H2SO4 


9 g HC2H3O2 = 9g / 325.3g/mole 

= 0.02767 moles of Pb(C2H3O2)2 


After reaction and mixture formation:
0.02767 moles of PbSO4 


0.05534 mole of HC2H3O2 


0.00000 mole of Pb(C2H3O2)2 


0.09176-0.02767= 0.06409 moles of H2SO4 


0.06409 moles x 98.08 g/mole = 6.29 g H2SO4

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 22nd, 2015

I need the answer in grams, you only gave H2SO4 in grams. I need all of them :)

Jul 22nd, 2015

okay let me do it for you. Don't worry

Jul 22nd, 2015

0.05534 mole of HC2H3O2 x60.052 g/mol =3.32 g of HC2H3O2

0.00000 mole of Pb(C2H3O2)2 = 0 g of Pb(C2H3O2)2

0.06409 moles x 98.08 g/mole = 6.29 g H2SO4

0.02767 moles of PbSO4  x 303.27g/mol = 8.39 g of PbSO4

Jul 22nd, 2015

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