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Multiple Choice
Propane (C₃H₈) reacts with oxygen in the air to produce carbon dioxide and water. In a particular experiment, 38.0 grams of carbon dioxide are produced from the reaction of 22.05 grams of propane with excess oxygen. What is the percentage yield in this reaction?
A
85.0%
B
90.0%
C
75.0%
D
95.0%
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1
Identify the balanced chemical equation for the combustion of propane: \( \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \).
Calculate the molar mass of propane (C₃H₈) by adding the atomic masses of its constituent atoms: \( 3 \times 12.01 \text{ g/mol (C)} + 8 \times 1.01 \text{ g/mol (H)} \).
Determine the moles of propane used in the reaction by dividing the given mass of propane (22.05 grams) by its molar mass.
Using the stoichiometry of the balanced equation, calculate the theoretical moles of carbon dioxide produced from the moles of propane. Since 1 mole of propane produces 3 moles of carbon dioxide, multiply the moles of propane by 3.
Convert the theoretical moles of carbon dioxide to grams using its molar mass (44.01 g/mol) and compare it to the actual mass of carbon dioxide produced (38.0 grams) to calculate the percentage yield using the formula: \( \text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \% \).