Question

When carbon disulfide burns in the presence of oxygen, sulfur dioxide and carbon dioxide are produced according to the following equation.

CS2 + 3O2 -----> CO2 + 2SO2
a. What is the theoretical yield of carbon dioxide if 2.50 mol of oxygen reacts with carbon disulfide?_________ g CO2 (answer should have 1 number after the decimal)
b. Calculate the percent yield of carbon dioxide if 32.4 g were produced. _______ % (answer should have 1 number after the decimal.

Answer 1

The theoretical yield of carbon dioxide when 2.50 mol of oxygen reacts with carbon disulfide is 36.7 g CO2. The percent yield of carbon dioxide is 88.3% if 32.4 g were actually produced.

Step-by-step explanation:

To calculate the theoretical yield of carbon dioxide (CO2) when 2.50 mol of oxygen (O2) reacts with carbon disulfide (CS2), we use the balanced chemical equation:

CS2 + 3O2 → CO2 + 2SO2

According to the balanced equation, 1 mole of CS2 reacts with 3 moles of O2 to produce 1 mole of CO2. Since oxygen is in excess, carbon disulfide is the limiting reactant. The molecular weight of CO2 is 44.01 g/mol.

As O2 is present in excess, we assume all of it reacts, and therefore the theoretical yield of CO2 is:

1. Calculate the moles of CS2 that would react with 2.50 mol O2: 2.50 mol O2 × (1 mol CS2/3 mol O2) = 0.833 mol CS2
2. Calculate the mass of CO2 produced: 0.833 mol CS2 × (1 mol CO2/1 mol CS2) × 44.01 g/mol = 36.7 g CO2

To calculate the percent yield of CO2 if 32.4 g were actually produced:

Percent Yield = (Actual Yield / Theoretical Yield) × 100%

= (32.4 g / 36.7 g) × 100%

= 88.3%