Calculate the mass of bromoethane required to obtain 20 g of ethanol at a theoretical yield of 75%.

Let’s implement the solution:
1. Let’s make the equation:
C2H5Br + KOH = C2H5OH + KBr – exchange, ethanol was formed;
2. Calculations by formulas:
M (C2H5Br) = 108.9 g / mol.
M (C2H5OH) = 46 g / mol.
3. Determine the mass, quantity of the product:
m (C2H5OH) = 20 / 0.75 = 26.6 g (theoretical weight).
Y (C2H5OH) = m / M = 26.6 / 46 = 0.58 mol.
Y (C2H5Br) = 0.58 mol since the amount of substances according to the equation is 1 mol.
4. Find the mass of the original substance:
m (C2H5Br) = Y * M = 0.58 * 108.9 = 63.2 g.
Answer: the mass of bromoethane is 63.2 g.