Calculate the mass of glucose required to obtain 276 g of ethyl alcohol.

1. Let us write down the equation for the reaction of alcoholic fermentation of glucose:

C6H12O6 → 2C2H5OH + 2CO2 ↑.

2. Let’s calculate the chemical amount of the obtained ethyl alcohol:

n (C2H5OH) = m (C2H5OH): M (C2H5OH);

M (C2H5OH) = 2 * 12 + 5 + 17 = 46 g / mol;

n (C2H5OH) = 276: 46 = 6 mol.

3. Determine the amount of initial glucose:

n (C6H12O6) = n (C2H5OH): 2 = 6: 2 = 3 mol.

4. Let’s calculate the mass of glucose:

m (C6H12O6) = n (C6H12O6) * M (C6H12O6);

M (C6H12O6) = 6 * 12 + 12 + 6 * 16 = 180 g / mol;

m (C6H12O6) = 3 * 180 = 540 g.

Answer: 540 BC