Methanol with an amount of a substance of 0.5 mol was heated with an excess of potassium bromide

Methanol with an amount of a substance of 0.5 mol was heated with an excess of potassium bromide and sulfuric acid and bromine-methane was obtained with a mass of 38 g. Determine the yield of bromomethane from the theoretically possible.

To solve the problem, we compose the equation of the process:

CH3OH + KBr (excess) = CH3Br + KOH – exchange, bromomethane is released;

Calculations:
M (CH3Br) = 92.9 g / mol;

Y (CH3OH) = 0.5 mol;

Y (CH3Br) = 0.5 mol since the amount of these substances is 1 mol.

We find the mass of the product, the calculation of the yield:
m (CH3Br) = Y * M = 0.5 * 92.9 = 46.45 g

W = m (practical) / m (theoretical) * 100;

W = 38 / 46.45 * 100 = 81.80%

Answer: the yield of bromomethane was 81.80%