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A 10 g piece of marble is placed in excess dilute HCl. When the reaction is complete, 1120 cm³ of CO2 is collected at STP. What is the percentage of CaCO3 in the marble?

1. 10%
2. 25%
3. 50%
4. 75%

Correct answer: 50%

Solution

CaCO3 + 2HCl -> CaCl2 + H2O + CO2. Moles of CO2 = 1120/22400 = 0.05 mol. Moles of CaCO3 = 0.05 mol. Mass of CaCO3 = 0.05 * 100 = 5 g. Percentage = (5/10)*100 = 50%.

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