What volume of carbon dioxide will be released during the interaction of 150 g of a 20% solution of acetic acid and CaCO3?

1. Let’s compose the equation of interaction of acetic acid and calcium carbonate:

2CH3COOH + CaCO3 → (CH3COO) 2Ca + H2O + CO2 ↑;

2.Calculate the mass of acetic acid:

m (CH3COOH) = w (CH3COOH) * m (solution) = 0.2 * 150 = 30 g;

3. find the chemical amount of acetic acid:

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

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

n (CH3COOH) = 30: 60 = 0.5 mol;

4.determine the amount of carbon dioxide:

n (CO2) = n (CH3COOH): 2 = 0.5: 2 = 0.25 mol;

5. find the volume of carbon dioxide:

V (CO2) = n (CO2) * Vm = 0.25 * 22.4 = 5.6 liters.

Answer: 5.6 liters.