A sample of and weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of mixture is

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Solution

Q.
A sample of $C a C O_{3}$ and $M g C O_{3}$ weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of mixture is

(Given molar mass in g $m o l^{- 1}$ $C a C O_{3} : 100, M g C O_{3} : 84$ ) [2024]

1 $1.023 g C a C O_{3} + 1.023 g M g C O_{3}$

2 $1.187 g C a C O_{3} + 1.187 g M g C O_{3}$

3 $1.023 g C a C O_{3} + 1.187 g M g C O_{3}$

4 $1.187 g C a C O_{3} + 1.023 g M g C O_{3}$

Ans.
(4)

   $Let mass of {CaCO}_{3} in the mixture i s x g, so mass of {MgCO}_{3} in the mixture is (2.21 - x) g .$

$Moles of {CaCO}_{3} (n_{{CaCO}_{3}}) = \frac{given mass}{molar mass} = \frac{x}{100} mol$

$Moles of {MgCO}_{3} (n_{{MgCO}_{3}}) = \frac{given mass}{molar mass} = \frac{2.21 - x}{84} mol$

When the mixture is heated carbon dioxide escapes out leaving solid mass containing CaO and MgO.

${CaCO}_{3} (s) \overset{Δ}{\to} C a O (s) + {CO}_{2} (g)$

${MgCO}_{3} (s) \overset{Δ}{\to} M g O (s) + {CO}_{2} (g)$

By stoichiometry:

   $Moles of CaO (n_{CaO}) = n_{{CaCO}_{3}} = \frac{x}{100} mol$

   $Moles of MgO (n_{MgO}) = n_{{MgCO}_{3}} = \frac{2.21 - x}{84} mol$

$Mass of CaO (W_{CaO}) = n_{CaO} \times molar mass = \frac{x}{100} \times 56 g$

$Mass of MgO (W_{MgO}) = n_{MgO} \times molar mass = \frac{2.21 - x}{84} \times 40 g$

   $Mass of residue = W_{CaO} + W_{MgO}$

$= \frac{x}{100} \times 56 + \frac{2.21 - x}{84} \times 40 = 1.152$

$\frac{56 x}{100} + \frac{(88.4 - 40 x)}{84} = 1.152$

   $\frac{56 \times 84 x + (88.4 - 40 x) 100}{8400} = 1.152$

   $704 x + 8840 = 9676.8$

$704 x = 836.8$

$x = 1.187$

   $Thus mass of {CaCO}_{3} in the mixture is 1.187 g and mass of {MgCO}_{3} in the mixture is (2.21 - 1.187) g = 1.023 g .$

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