Calculate the mass of bromoethane obtained by the reaction of 92 g of ethanol with an excess of hydrogen bromide

Calculate the mass of bromoethane obtained by the reaction of 92 g of ethanol with an excess of hydrogen bromide, if the proportion of the product yield is 85% of the theoretically possible.

To solve, we write the equation:
С2Н5ОН + НBr = C2H5Br + H2O – exchange reaction, bromoethane was obtained;
Let’s calculate the molar masses of substances:
M (C2H5OH) = 46 g / mol;
M (C2H5Br) = 108.9 g / mol.
Let’s calculate the number of moles of ethanol, if the mass is known:
Y (C2H5OH) = m / M = 92/46 = 2 mol;
Y (C2H5Br) = 2 mol, since the number of moles of these substances according to the equation is equal to 1 mol.
Let us determine the theoretical mass of bromoethane:
m (C2H5Br) = Y * M = 2 * 108.9 = 217.8 g.
We find a practical mass by the formula:
W = m (practical) / m (theoretical) * 100;
m (practical) = 0.85 * 217.8 = 185.13 g.
Answer: the mass of bromoethane is 185.13 g.