We’re being asked to calculate the **percent yield of CH _{3}Br** in the reaction of

CH_{4}O + HBr → CH_{3}Br + H_{2}O

Recall that ** percent yield** is given by:

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

We know the reaction produced 12.23 g CH_{3}Br; this is the actual yield of the reaction. To calculate the theoretical yield, we need to do the following:

Mass of CH_{4}O **(molar mass of CH _{4}O) **→ Moles of CH

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

What is the percent yield for CH_{3}Br if 12.23 g is produced when 5.00 g of CH_{4}O reacts with excess HBr? CH_{4}O + HBr → CH_{3}Br + H_{2}O

A. 16%

B. 67%

C. 84%

D. 82.6%

E. 12%

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