What mass of glucose must be fermented to obtain ethyl alcohol weighing 46 grams, if the yield is practically 90%

To solve the problem, we compose the equation:
X g -? m = 46 g; W = 90?
1. С6Н12О6 = 2СО2 + 2С2Н5ОН – alcoholic fermentation, ethanol was obtained;
2. Calculations by formulas:
M (C6H12O6) = 180 g / mol;
M (C2H5OH) = 46 g / mol.
3. Determine the mass of alcohol, its amount:
m (theoretical) = 0.90 * 46 = 41.4 g;
Y (C2H5OH) = m / M = 41.4 / 46 = 0.9 mol.
4. Proportion:
X mol (C6H12O6) – 0.9 mol (C2H5OH);
– 1 mol – 2 mol from here, X mol (C6H12O6) = 1 * 0.9 / 2 = 0.45 mol.
5. Find the mass of the original substance:
m (C6H12O6) = Y * M = 0.45 * 180 = 81 g.
Answer: for fermentation, you need glucose weighing 81 g.