20.0g 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?
(At. Wt. Mg = 24)

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Solution

The correct option is B 84
Percentage purity = mass of $M g C O_{3}$ actually present / mass of $M g C O_{3}$ that should be theoretically present.
The denominator is 20g i n this case.

$M g C O_{3} \to M g O + C O_{2}$

8 g of $M g O$ formed - 8/40 = 0.2 mol.
Hence, 0.2 mol of $M g C O_{3}$ reacted.

% purity = $0.2 \times (24 + 12 + 48) / 20$ = 84 %

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Similar questions

$20.0 g$ of a magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0 g$ of magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. Wt. $M g = 24$ )

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

M g = 24

]

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20.0g 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? (At. Wt. Mg = 24)

The correct option is B 84Percentage purity = mass of MgCO3 actually present / mass of MgCO3 that should be theoretically present. The denominator is 20g i n this case. MgCO3→MgO + CO2 8 g of MgO formed - 8/40 = 0.2 mol. Hence, 0.2 mol of MgCO3 reacted. % purity = 0.2×(24+12+48)/20 = 84 %

20.0g 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?
(At. Wt. Mg = 24)