How much hydrogen can be released when calcium interacts with 22.5 g of aminoacetic acid?

1.2NH2CH2COOH + Ca → (NH2CH2COO) 2Ca + H2 ↑;

2.n (NH2CH2COOH) = m (NH2CH2COOH): M (NH2CH2COOH);

M (NH2CH2COOH) = 14 + 1 * 5 + 12 * 2 + 16 * 2 = 67 g / mol;

n (NH2CH2COOH) = 22.5: 67 = 0.3358 mol;

3. according to the reaction equation, the amount of hydrogen is half the amount of aminoacetic acid:

n (H2) = n (NH2CH2COOH): 2;

n (H2) = 0.3358: 2 = 0.1679 mol;

4.Calculate the volume of hydrogen using the formula for gases under normal conditions:

V (H2) = n (H2) * Vm

V (H2) = 0.1679 * 22.4 = 3.76 dm3.

Answer: 3.76 dm3.