Methanol (CH₄O) is converted to bromomethane (CH₃Br) as follows:
CH₄O + HBr $\to$ CH₃Br + H₂O
If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?

Question

Methanol (CH₄O) is converted to bromomethane (CH₃Br) as follows:
CH₄O + HBr

\to

CH₃Br + H₂O
If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?

Quizplus · Accepted Answer

To calculate the theoretical yield, we first need to find the limiting reagent. The balanced chemical equation shows us that the mole ratio of methanol to bromomethane is 1:1, so we can use the number of moles of methanol to find the theoretical yield. 5.00 g CH₄O × (1 mol CH₄O/32.04 g CH₄O) × (1 mol CH₃Br/1 mol CH₄O) × (94.94 g CH₃Br/1 mol CH₃Br) = 14.84 g CH₃Br Therefore, the theoretical yield is 14.84 g CH₃Br. % Yield = (Actual Yield/Theoretical Yield) × 100 % Yield = (12.23 g/14.84 g) × 100 % Yield = 82.6% Therefore, the answer is B, 82.6%.

Methanol (CH4O) Is Converted to Bromomethane (CH3Br) as Follows: →\to→

Methanol (CH₄O) Is Converted to Bromomethane (CH₃Br) as Follows:
$\to$