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A 62.5 g mixture of CaCO3 and SiO2 is treated with excess hydrochloric acid. Only CaCO3 reacts: CaCO3 + 2HCl -> CaCl2 + CO2 + H2O. The mass of CO2 produced is 1.1 g. What is the mass percentage of CaCO3 in the original mixture?

1. 4%
2. 8%
3. 10%
4. 2.5%

Correct answer: 4%

Solution

From 1.1 g CO2, moles CO2 = 1.1/44 = 0.025 mol. Since CaCO3 and CO2 are in a 1:1 ratio, mass CaCO3 = 0.025 * 100 = 2.5 g. Mass% = (2.5/62.5) * 100 = 4%.

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