Final answer:
The percent yield of the combustion reaction where 44.03 g of butane produces 84.63 g of CO2 is approximately 63.5%, calculated by comparing the actual yield to the theoretical yield obtained from stoichiometry.
Step-by-step explanation:
To determine the percent yield of the reaction where 44.03 g of C4H10 (butane) combusts to produce 84.63 g of CO2 (carbon dioxide), we first must calculate the theoretical yield of CO2 using the stoichiometry of the balanced chemical equation for the combustion of butane:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
From the balanced equation, we see that 2 moles of C4H10 produce 8 moles of CO2. The molar mass of C4H10 is 58.12 g/mol, so 44.03 g of C4H10 is equal to 44.03 g / 58.12 g/mol = 0.757 moles of C4H10. Since 2 moles of C4H10 produce 8 moles of CO2, 0.757 moles of C4H10 would ideally produce 0.757 moles C4H10 / 2 moles C4H10 × 8 moles CO2 = 3.028 moles of CO2, where each mole of CO2 has a molar mass of 44.01 g/mol, leading to a theoretical yield of 3.028 moles CO2 × 44.01 g/mol = 133.24 g CO2.
The percent yield can then be calculated using the actual yield (84.63 g) and the theoretical yield (133.24 g) with the following formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100 = (84.63 g / 133.24 g) × 100
Thus, the percent yield = (84.63 / 133.24) × 100 ≈ 63.5%