90 g of glucose was subjected to alcoholic fermentation, the alcohol yield was 80%

90 g of glucose was subjected to alcoholic fermentation, the alcohol yield was 80% of the theoretically possible. Find the mass of the obtained alcohol.

Let’s write down the reaction scheme and arrange the coefficients:
C6H12O6 = 2C2H5OH + 2CO2
It can be seen from the reaction equation that from 1 mol of glucose 2 mol of alcohol is formed.
Let’s calculate the amount of substances n alcohol and glucose:
n (ch.) = m (ch.) * M (ch.) = 90/180 = 0.5 mol.
n (cn.) = 2 * n (ch.) = 1 mol.
Let’s calculate the theoretical mass of alcohol m (cn. Theory) = n (cn.) * M (cn.) = 1 * 46 = 46 g.
Taking into account the output w, the practical mass of alcohol m (c.p.) = m (c.p.theor.) * W = 46 * 0.8 = 36.8 g.