Answer:
sulfuric acid: 5.24 g
lead(II) acetate: 0 g
lead(II) sulfate: 6.99 g
acetic acid: 2.77 g
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.
= 25 g solution