When 60.0 g of CH₄ reacts with excess O₂, the actual yield of CO₂ is 112 g. What is the percent yield? CH₄ + 2O₂ → CO₂ + 2H₂O

Question

Quizplus · Accepted Answer

1 mole of CH4 produces 1 mole of COS. The molar mass of CH4 is 16 g/mol and the molar mass of COS is 60 g/mol. Using stoichiometry, we can find that the theoretical yield of COS is:

(60.0 g CH4) / (16.0 g/mol CH4) x (1 mol COS / 1 mol CH4) x (60.0 g/mol COS) = 225 g

Percent yield is actual yield divided by theoretical yield, multiplied by 100%:

Percent yield = (112 g / 225 g) x 100% = 49.8%

Therefore, the closest choice is C, 67.9%, but it is still incorrect. The percent yield should be rounded to one decimal place, giving a percent yield of 49.8%.

When 600 G of CH₄ Reacts with Excess O2, the Actual

When 600 G of CH₄ Reacts with Excess O₂, the Actual