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What is the purity percentage of MgCO3 if 20 g of MgCO3 forms 9.52 g of MgO?

1. 80%
2. 84%
3. 90%
4. 92%

Correct answer: 84%

Solution

The molar mass of MgCO3 is 84 g/mol, and it decomposes to form MgO (molar mass 40 g/mol). From stoichiometry, 84 g of MgCO3 gives 40 g of MgO. For 9.52 g of MgO, the corresponding MgCO3 is (9.52 × 84) / 40 = 20 g. Since the entire 20 g of the sample is MgCO3, the purity is 100 × (20/20) = 84%.

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