How much ethanol can be obtained by fermenting 162 g of glucose?

Ethanol (ethyl alcohol) can be obtained by the glucose fermentation reaction:
C6H12O6 = 2 C2H5OH + 2 CO2.
M (C6H12O6) = 12×6 + 1×12 + 16x 6 = 180g / mol, since 1 mol enters the reaction, then m (C6H12O6) = 180g;
M (C2H5OH) = 12×2 + 1×6 +16 = 46g / mol, since according to the equation 2 mol of alcohol is formed,
then m (C2H5OH) = 46×2 = 92g
If from 180 g of glucose 92 g of alcohol is formed, then from 162 g – xg.
Let’s compose and solve the proportion:
180/162 = 92 / x, x = 162×92 / 180 = 82.2
Answer: when fermenting 162 g of glucose, 82.2 g of ethanol will be released.