Question

Methanol (ch3oh) is converted to bromomethane (ch3br) as follows: ch3oh + hbr 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?

Answer 1

Answer:

82.39% is the percentage yield.

Explanation:

$CH_3OH + HBr\rightarrow CH_3Br+ H_2O$

Moles of methanol = $\frac{5.00 g}{32 g/mol}=0.15625 mol$

According to reaction, 1 mol of methanol gives 1 mole of bromo methane .

Then 0.15625 moles of methanol will gives:

$\frac{1}{1}\times 0.15625 mol=0.15625 mol$ of bromomethane

Mass of 0.15625 moles of bromomethane =

$0.15625 mol\times 95 g/mol=14.84375 g$

Theoretical yield = 14.84375 g

Experimental yield = 12.23 g

$\%yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\% yield=\frac{12.23 g}{14.84375 g}\times 100=82.39\%$

82.39% is the percentage yield.