$20.0 g$ 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 ?
(At. Wt. : $M g$ = 24)

60

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84

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75

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96

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Solution

The correct option is B $84$
$M g C O_{3} (s) \to M g O (s) + C O_{2} (g)$
Moles of $M g C O_{3} = \frac{20}{84} mol$
From the above equation
1 mole of $M g C O_{3}$ gives 1 mole $M g O$
$\frac{20}{84} \times 40 g = \frac{200}{21} g M g O$
Practical yield of $M g O = 8 g M g O$
$∴$ % purity = $\frac{8 \times 21}{200} \times 100 = 84 %$

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

The correct option is B 84MgCO3(s)→MgO(s)+CO2(g) Moles of MgCO3=2084 mol From the above equation 1 mole of MgCO3 gives 1 mole MgO 2084×40 g=20021 g MgO Practical yield of MgO=8 gMgO ∴ % purity =8×21200×100=84%

$20.0 g$ 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 ?
(At. Wt. : $M g$ = 24)