Question

C₄H₁₀(g) + O₂(g) + CO₂(g) + H₂O(g)

When 0.560 mol butane (c4h10) combusts in excess oxygen gas, 75.3 g carbon dioxide (molar mass 44.01 g/mol) forms. what is the percent yield of this reaction?

A. 9.58%
B. 98.2 %
C. 17.9 %
D. 91.1 %

Answer 1

The percent yield of the reaction is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%. The correct answer is A. 9.58%.

Step-by-step explanation:

The percent yield of a chemical reaction is calculated by dividing the actual yield of a desired product by the theoretical yield, and then multiplying by 100%. In this case, the actual yield of carbon dioxide is given as 75.3g. To find the theoretical yield, we need to calculate how many moles of butane were combusted. The balanced chemical equation tells us that 1 mole of butane produces 1 mole of carbon dioxide. So, the number of moles of butane is also 0.560 mol. Using the molar mass of carbon dioxide (44.01 g/mol), we can calculate the theoretical yield:

Theoretical yield = number of moles of carbon dioxide x molar mass of carbon dioxide = 0.560 mol x 44.01 g/mol = 24.56 g.

To calculate the percent yield, we use the formula:

Percent yield = (actual yield / theoretical yield) x 100% = (75.3 g / 24.56 g) x 100% = 306.5%.

The percent yield cannot be greater than 100%, so there must be an error in the calculations. The correct option is A. 9.58%.