20.0 grams of Magnesium carbonate sample decomposes on heating to give CO2 and 8.0 g magnesium oxide. What will be the percentage purity of Magnesium carbonate in the sample

MgCO3⏟1mole→Mgo⏟1mole+CO2 84 g of MgCO3 gives 40 g of MgO∵20 g of MgCO3 →?=40×2084=9.52g of MgOSince 8.0 g of MgO is formed, purity of sample =89.52×100=84%

20.0 grams of Magnesium carbonate sample decomposes on heating to give CO2 and 8.0 g magnesium oxide. What will be the percentage purity of Magnesium carbonate in the sample

MgCO3⏟1mole→Mgo⏟1mole+CO2 84 g of MgCO3 gives 40 g of MgO∵20 g of MgCO3 →?=40×2084=9.52g of MgOSince 8.0 g of MgO is formed, purity of sample =89.52×100=84%

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20.0 grams of Magnesium carbonate sample decomposes on heating to give CO₂ and 8.0 g magnesium oxide. What will be the percentage purity of Magnesium carbonate in the sample

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Detailed Solution

$\underset{1 mole}{\underset{⏟}{{MgCO}_{3}}} \to \underset{1 mole}{\underset{⏟}{Mgo}} + {CO}_{2}$
84 g of MgCO₃ gives 40 g of MgO
$∵$ 20 g of MgCO₃→?
$= \frac{40 \times 20}{84} = 9 .52 g of MgO$
Since 8.0 g of MgO is formed, purity of sample $= \frac{8}{9 .52} \times 100 = 84 %$