20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?

(Atomic weight: Mg = 24)NEET 2015 (Phase 2)

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M g C O 3 s ⟶ M g O s + C O 2 g M o l e s o f M g C O 3 = 20 84 = 0.238 m o l From above equation. 1 mole M g C O 3 gives 1 mole MgO ∴ 0.238 mole M g C O 3 will give 0.238 mole MgO = 0.238 × 40 g = 9.523 g M g O P r a c t i c a l y i e l d o f M g O = 8 g M g O ∴ % P u r i t y = 8 9.523 × 100 = 84 %

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