Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and a solution of acetic acid.

Question

If 10.0g of sulfuric acid and 10.0g of lead (II) acetate are mixed, calculate the number of grams of sulfuric acid, lead (II) acetate, lead (II) sulfate, and acetic acid present after the reaction is complete. Show all of your work.

Answer 1

Let's take a look at the equation for the reaction taking place:

H₂SO₄(aq) + Pb(C₂H₃O₂^-)2(aq) ==> PbSO₄(s) + 2CH₃COOH

From here on, I will use Ac to represent the acetate anion (C₂H₃O₂^-) and HAc to represent acetic acid.

First, we need to find out which reactant is present in limiting supply:

moles H2SO4 present = 10.0 g H2SO4 x 1 mol/98 g = 0.102 moles H2SO4

moles Pb(Ac)₂ present = 10.0 g Pb(Ac)₂ x 1 mol/325 g = 0.0307 moles Pb(Ac)₂

So, clearly Pb(Ac)₂ is limiting and as such will dictate how much PbSO4 and HAc can be formed. All of the Pb(Ac)₂ will be used up, and since H2SO4 is in excess, there will be some left over. The calculations follow:

Using dimensional analysis...

0.0307 moles Pb(Ac)₂ x 1 mol PbSO4/mol Pb(Ac)₂ = 0.0307 moles PbSO4 formed

grams PbSO4 = 0.0307 moles x 303 g/mol = 9.32 g PbSO4

molesH2SO4 left over = 0.102 mol H2SO4 - 0.0307 moles used = 0.0713 moles H2SO4 left over

grams H2SO4 left over = 0.0713 moles x 98 g/mol = 6.99 g H2SO4

moles Pb(Ac)₂ left over = 0

grams Pb(Ac)₂ left over = 0 g Pb(Ac)₂

moles HAc formed = 0.0307 moles Pb(Ac)₂ x 2 mol HAc/ mol Pb(Ac)₂ = 0.0614 moles HAc formed

grams HAc formed = 0.0614 moles HAc x 60 g/mol = 3.68 g HAc