When ethanol interacted with sodium, hydrogen was released with a volume of 0.336 dm3

When ethanol interacted with sodium, hydrogen was released with a volume of 0.336 dm3. Determine the mass of the reacted alcohol.

1. We write down the reaction equation:

2C2H5OH + 2Na → 2C2H5ONa + H2 ↑;

2. find the chemical amount of released hydrogen:

n (H2) = V (H2): Vm = 0.336: 22.4 = 0.015 mol;

3. Determine the amount and mass of the alcohol reacted:

n (C2H5OH) = n (H2) * 2 = 0.015 * 2 = 0.03 mol;

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

M (C2H5OH) = 12 * 2 + 5 + 16 + 1 = 46 g / mol;

m (C2H5OH) = 0.03 * 46 = 1.38 g.

Answer: 1.38 g.