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

label Chemistry
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

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

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Jul 22nd, 2015
...
Jul 22nd, 2015
Oct 18th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer