Question

1. For the following reaction, 5.94 g of water are mixed with excess sulfur dioxide. the reaction yield 21.3 g of sulfurous acid. what is the theoretical yoeld of sulfurous acid in grams? what is the percent yield for this reaction?

2. For the following reaction 3.02 g of water are mixed with excess phosphorus pentoxide. The reaction yields 7.19 g of phosphoric acid. what is the theoretical yield of phosphoric acid in grams? what is the percent yield for this reaction?

3. For the following reaction 3.79 g of sulfur are mixed with excess carbon monoxide. the reaction yields 5.63 g of sulfur dioxide. what is the theoretical yield of sulfur dioxide? what is the percent yield of sulfur dioxide?

Answer 1

Molar mass of sulfur dioxide = 64.066 g/mol

Molar mass of water = 18.015 g/mol

Molar mass of sulfurous acid = 82.08 g/mol

Molar mass of phosphorus pentoxide= 283.886 g/mol

Molar mass of phosphoric acid = 97.994 g/mol

Molar mass of carbon monoxide = 28.01 g/mol

Molar mass of sulfur= 32.065 g/mol

Molar mass of sulfur dioxide = 64.066 g/mol

1.

SO₂ + H₂O --> H₂SO₃

Mass of water mixed = 5.94 g

Moles of water mixed = 5.95 g / 18.015 g/mol = 0.33

By stoichiometry, theoretical moles of sulfurous acid produced = 0.33 * 1/1 = 0.33

Theoretical yield of sulfurous acid produced = 0.33 mol *82.08 g/mol = 27.1 g

% yield of reaction = actual yield / theoretical yield * 100

= 21.3 / 27.1 * 100 = 78.57 %

2.

P₄O₁₀ + 6 H₂O --> 4 H₃PO₄

Mass of water mixed = 3.02 g

Moles of water mixed = 3.02 g / 18.015 g/mol = 0.168

By stoichiometry, theoretical moles of phosphoric acid produced = 0.168 * 4/6 = 0.11

Theoretical yield of phosphoric acid produced = 0.11 mol *97.994 g/mol = 10.95 g

% yield of reaction = actual yield / theoretical yield * 100

= 7.19 / 10.95 * 100 = 65.65 %

3.

2 CO + S --> SO2 + 2 C

Mass of sulfur mixed = 3.79 g

Moles of sulfur mixed = 3.79 g / 32.065 g/mol = 0.118

By stoichiometry, theoretical moles of sulfur dioxide produced = 0.118 * 1/1 = 0.118

Theoretical yield of sulfur dioxide produced = 0.118 mol * 64.066 g/mol = 7.57 g

% yield of reaction = actual yield / theoretical yield * 100

= 5.63 / 7.57 * 100 = 74.35 %