# Determine the volume of gas that will be released when 200 g of 6.9% potassium

September 30, 2021 | education

| **Determine the volume of gas that will be released when 200 g of 6.9% potassium carbonate solution interacts with an excess of hydrochloric acid.**

1. Let’s compose the reaction equation:

CaCO3 + 2HCl = CaCl2 + H2O + CO2 ↑;

2. find the mass of calcium carbonate:

m (CaCO3) = w (CaCO3) * m (solution);

m (CaCO3) = 0.069 * 200 = 13.8 g;

3.Calculate the chemical amount of carbonate:

n (CaCO3) = m (CaCO3): M (CaCO3);

M (CaCO3) = 40 + 12 + 3 * 16 = 100 g / mol;

n (CaCO3) = 13.8: 100 = 0.138 mol;

4. find the chemical amount and volume of carbon dioxide released:

n (CO2) = n (CaCO3) = 0.138 mol;

V (CO2) = n (CO2) * Vm = 0.138 * 22.4 = 3.0912 dm3.

Answer: 3.0912 dm3.

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