18.4g mixture of $C a C O_{3}$ and $M g C O_{3}$ , on strongly heating gives 8.8g of $C O_{2}$ gas. Calculate mole percentage of $M g C O_{3}$ in the mixture.

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Solution

Let $C a C O_{3} b e x g$ .
Then, $M g C O_{3} i s (18.4 - x) g$ .
$C a C O_{3} \to C a O + C O_{2}$
1 mol 1 mol 1 mol
100g 56g 44g
xg $\frac{44 x}{100}$ g
$M g C O_{3} \to M g O + C O_{2}$
1 mol 1 mol 1 mol
84g 40g 44g
(18.4-x)g $\frac{44 (18.4 - x)}{84} g$
Total amount of $C O_{2} = \frac{44 x}{100} + \frac{44 (18.4 - x)}{84}$ = 8.8 (given).
This gives x = 10g.
$∴ C a C O_{3} = 10 g = 0.1 m o l$ .
$M g C O_{3} = 18.4 - 10 = 8.4 g = 0.1 m o l$ .
Total moles of $C a C O_{3}$ and $M g C O_{3} = 0.2 m o l$ .
$∴ C a C O_{3} = 50 m o l %$ .
$M g C O_{3} = 50 m o l %$ .

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18.4g mixture of CaCO3 and MgCO3, on strongly heating gives 8.8g of CO2 gas. Calculate mole percentage of MgCO3 in the mixture.

18.4g mixture of $C a C O_{3}$ and $M g C O_{3}$ , on strongly heating gives 8.8g of $C O_{2}$ gas. Calculate mole percentage of $M g C O_{3}$ in the mixture.