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

[Atomic weight of Mg=24]

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\\M g C O_{3} \rightarrow M g O+C O_{2}$ \\\\ The molar masses of $M g C O_{3}$ and $M g O$ are $24+12+48=84 \mathrm{g} / \mathrm{mol}$ and $24+16=40 \mathrm{g} / \mathrm{mol}$ respectively. \\\\ Thus, $8 \ \mathrm{g} \ \mathrm{MgO}$ will be obtained from $\frac{84}{40} \times 8=16.8 \mathrm{mg}$ of $\mathrm{MgCO}_{3}$ \\\\ The percentage purity of magnesium carbonate in the sample is $\frac{16.8}{20} \times 100=80 \%$

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