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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