Find V (CO2) that will be released when hydrochloric acid interacts with 28 g of calcium carbonate having 15% impurities.

univerkov | January 7, 2021 | education

We are told that we have 28 grams of calcium carbonate from 15% impurities. So the percentage of pure calcium carbonate is: 100% – 15% = 85%
The mass of pure calcium carbonate is: m = 28 g * 0.85 = 23.8 g
We write the equation:
2HCl + CaCO3 = CaCl2 + CO2 + H2O
23.8 – x
100 – 22.4
Then V (CO2) = 23.8 * 22.4 / 100 = 5.33 L

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