Question

If the combustion of 44.23 g of C₄H₁₀ produces 108.75 g of CO₂, what is the percent yield of the reaction? (Assume oxygen is in excess.)

Answer 1

The percent yield of the reaction is 81.23%.

Step-by-step explanation:

To calculate the percent yield of a reaction, you need to compare the actual yield (the amount of product produced in the experiment) with the theoretical yield (the amount of product predicted by stoichiometry). In this case, the combustion of 44.23 g of C₄H₁₀ produces 108.75 g of CO₂. The balanced equation for this reaction is:

C₄H₁₀ + 6.5O₂ → 4CO₂ + 5H₂O

From the balanced equation, we can see that for every 1 mole of C₄H₁₀ combusted, we should get 4 moles of CO₂. Firstly, we need to calculate the theoretical yield of CO₂:

Molar mass of C₄H₁₀ = 58.12 g/mol

Molar mass of CO₂ = 44.01 g/mol

Moles of C₄H₁₀ = 44.23 g / 58.12 g/mol = 0.761 mol

Therefore, the theoretical yield of CO₂ is:

Theoretical yield of CO₂ = 0.761 mol * 4 mol CO₂/1 mol C₄H₁₀ * 44.01 g/mol = 133.81 g

The percent yield can then be calculated as:

Percent yield = (Actual yield / Theoretical yield) * 100

Percent yield = (108.75 g / 133.81 g) * 100 = 81.23%