Calculate the volume of hydrogen released during the interaction of acetic acid with a mass of 12 g

Calculate the volume of hydrogen released during the interaction of acetic acid with a mass of 12 g with magnesium 4.8, the product yield is 90%.

1. The interaction of acetic acid with magnesium proceeds according to the equation:

2CH3COOH + Mg → (CH3COO) 2Mg + H2 ↑;

2.Calculate the chemical amounts of acid and magnesium:

n (CH3COOH) = m (CH3COOH): M (CH3COOH);

M (CH3COOH) = 12 + 3 + 12 + 32 + 1 = 60 g / mol;

n (CH3COOH) = 12: 60 = 0.2 mol;

n (Mg) = m (Mg): M (Mg) = 4.8: 24 = 0.2 mol;

3. acetic acid is taken in deficiency, using its amount, we determine the theoretical amount of released hydrogen:

ntheor (H2) = n (CH3COOH): 2 = 0.2: 2 = 0.1 mol;

4. find the practical chemical amount of hydrogen:

npract (H2) = ntheor (H2) * ν = 0.1 * 0.9 = 0.09 mol;

5.Calculate the volume of gas:

V (H2) = npract (H2) * Vm = 0.09 * 22.4 = 2.016 l.

Answer: 2.016 liters.