Question

Sulfur dioxide gas (SO2) reacts with excess oxygen gas (O2) and excess liquid water (H2O) to form liquid sulfuric acid (H2SO4). In the laboratory, a chemist carries out this reaction with 67.2 L of sulfur dioxide and gets 250 g of sulfuric acid.

• Calculate the theoretical yield of sulfuric acid.

• Calculate the percent yield of the reaction.
(One mole of any gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those conditions for this question.)

Answer 1

1) Answer is: the percent yield of the reaction is 84.5%.

Balanced chemical reaction: 2SO₂ + 2H₂O + O₂ → 2H₂SO₄.

V(SO₂) = 67.2 L; voulume of sulfur dioxide.

Vm = 22.4 L/mol; molar volume.

m(H₂SO₄) = 250 g; actual mass of sulfuric acid.

n(SO₂) = V(SO₂) ÷ Vm.

n(SO₂) = 67.2 L ÷ 22.4 L/mol.

n(SO₂) = 3 mol; amount of sulfur dioxide.

2) From balanced chemical reaction: n(SO₂) : n(H₂SO₄) = 2 : 2 (1 : 1).

n(H₂SO₄) = n(SO₂).

n(H₂SO₄) = 3 mol; amoun of sulfuric acid.

M(H₂SO₄) = 98.08 g/mol; molar mass of sulfuric acid.

m(H₂SO₄) = n(H₂SO₄) · M(H₂SO₄).

m(H₂SO₄) = 3 mol · 98.08 g/mol.

m(H₂SO₄) = 294.24 g; theoretical mass of sulfuric acid.

Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.

the percent yield = actual yield / theoretical yield.

the percent yield = 250 g ÷ 294.24 g · 100%.

the percent yield = 84.5 %.