Consider the following balanced equation for the following reaction:

15O2 (g) + 2C6H5COOH (aq) → 14CO2 (g) + 6H2O (l)

Determine the amount of CO2 (g) formed in the reaction if the percent yield of CO2 (g) is 83.0% and the theoretical yield of CO2 (g) is 1.30 moles.

The actual yield of CO2 is 1.079 moles or 47.5 grams formed

Explanation:

Step 1: Data given

Number of CO2 = 1.30 moles

Percent yield = 83.0 %

Step 2: The balanced equation

15O2 (g) + 2C6H5COOH (aq) → 14CO2 (g) + 6H2O (l)

Step 3: Calculate the number moles of CO2 formed

1.30 moles CO2 = 100 %

The actual amount of moles = 0.83 * 1.30 = 1.079 moles

Step 4: Calculate the percent yield of the reaction

We can control this by calculating the percent yield of the reaction

% yield = (actual yield / theoretical yield) * 100 %

% yield = (1.079 moles / 1.30 moles) * 100 %

% yield = 83.0 %

Step 5: Calculate thr mass of CO2 produced

Mass CO2 = moles * molar mass CO2

Mass CO2 = 1.079 moles * 44.01 g/mol

Mass CO2 = 47.5 grams

The actual yield of CO2 is 1.079 moles or 47.5 grams formed