Question

Solutions of sulfuric acid and lead(ll) acetate react to form solid lead(ll) sulfate and a solution of acetic acid. If 7.50 g ofsulfuric acid and 7.50 g of lead(ll) acetate are mixed, calculate the number of grams of sulfuric acid, Iead(ll) acetate, lead(ll)sulfate, and acetic acid present in the mixture after the reaction is complete.

Answer 1

sulfuric acid: 5.24 g

lead(II) acetate: 0 g

lead(II) sulfate: 6.99 g

acetic acid: 2.77 g

Step-by-step explanation:

The molecular formula of the compounds and their molar masses are:

Sulfuric acid: H2SO4, 98 g/mol;

Lead(II) acetate: Pb(Ac)2, 325 g/mol (Ac = C2H3O2);

Lead(II) sulfate: PbSO4, 303 g/mol;

Acetic acid: HAc, 60 g/mol

Thus, the reaction is:

H2SO4 + Pb(Ac)2 → PbSO4 + 2HAc

The stoichiometry of the reaction is 1:1:1:2. Let's first identify which of the reactant is the limiting, the one that will be totally consumed:

1 mol H2SO4 ------- 1 mol Pb(Ac)2

Transforming the relation to mass, multiplying the number of moles by the molar mass, and assuming H2SO4 as the liting reactant:

98 g H2SO4 ------------ 325 g Pb(Ac)2

7.50 g ------------ x

By a simple direct three rule:

98x = 2437.5

x = 24.87 g of Pb(Ac)2

Thus, there are fewer Pb(Ac)2 than is necessary, so it's the limiting.

So, the mass of sulfuric acid that reacts is:

325 g Pb(Ac)2 ---------------- 98 g H2SO4

7.50g ---------------- x

By a simple direct three rule:

325x = 735

x = 2.26 g of H2SO4

So, the left mass is 7.50 - 2.26 = 5.24 g.

The lead(II) acetate reacts completely, thus the final mass is 0.

The stoichiometry with lead (II) sulfate is:

1 mol Pb(Ac)2 ---------- 1 mol PbSO4

Passing by mass:

325 g Pb(Ac)2 --------------- 303 g PbSO4

7.5 g --------------- x

By a simple direct three rule:

325x = 227.25

x = 6.99 g formed of PbSO4

And the stoichiometry with HAc is:

1 mol Pb(Ac)2 ------- 2 moles HAc

Passing by mass

325 g Pb(Ac)2 ----- 120 g HAc

7.50 g ------ x

By a simple direct three rule:

325x = 900

x = 2.77 g of HAc.