Determine the mass of alcohol that was obtained by fermentation of 55g of glucose containing 8% impurities

1) Let’s write the reaction equation:

С6H12O6 → 2C2H5OH + 2CO2

2) Find the amount of glucose substance:

m (C6H12O6) = m (glucose with impurities) * ω (pure glucose) = 55 * (1 – 0.08) = 55 * 0.92 = 50.6 g;

n (C6H12O6) = m (C6H12O6) / Mr (C6H12O6) = 50.6 / 180 = 0.281 mol;

3) Find the mass of alcohol:

n (C2H5OH) = 2n (C6H12O6) = 2 * 0.281 = 0.562 mol;

m (C2H5OH) = n (C2H5OH) * Mr (C2H5OH) = 0.562 * 46 = 25.825 g.

Answer: 25.825 g.