Solution: What is the percent yield for CH₃Br if 12.23 g is produced when 5.00 g of CH₄O reacts with excess HBr? CH₄O + HBr → CH₃Br + H₂OA. 16% B. 67% C. 84% D. 82.6% E. 12%

We’re being asked to calculate the percent yield of CH₃Br in the reaction of 5.00 g HBr with excess CH₄O. The balanced chemical equation is:

 CH₄O + HBr → CH₃Br + H₂O

Recall that percent yield is given by:

$\boxed{\mathbf{\%}\mathbf{yield}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{actual}\mathbf{ }\mathbf{yield}}{\mathbf{theoretical}\mathbf{ }\mathbf{yield}}\mathbf{x}\mathbf{100}}$

We know the reaction produced 12.23 g CH₃Br; this is the actual yield of the reaction. To calculate the theoretical yield, we need to do the following:

Mass of CH₄O (molar mass of CH₄O) → Moles of CH₄O (mole-to-mole comparison) → Moles of CH₃Br (molar mass of CH₃Br) → Mass of CH₃Br

Since HBr is in excess, we can simply ignore it in our calculations.