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

User Generated

Thppvznlznl

Science

Description

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

User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Explanation & Answer

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

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)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**


Anonymous
Super useful! Studypool never disappoints.

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4

Related Tags