1.84 g of a mixture of $C a C O_{3}$ and $M g C O_{3}$ was heated to a constant weight. The constant weight of the residue was found to be 0.96 g. calculate the percentage composition of the mixture.. $(C a = 40, M g = 24, C = 12, O = 16)$

C a C O_{3} = 54.34 %, M g C O_{3} = 45.66 %

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M g C O_{3} = 54.34 %, C a C O_{3} = 45.66 %

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C a C O_{3} = 45.11 %, M g C O_{3} = 54.89 %

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M g C O_{3} = 73.24 %, C a C O_{3} = 26.76 %

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Solution

The correct option is C $C a C O_{3} = 45.11 %, M g C O_{3} = 54.89 %$
We have,
$C a C O_{3} + M g C O_{3} \to C a O + M g O + C O_{2} ↑$

Mass of calcium carbonate = x g
Mass of magnesium carbonate = (1.84 - x) g
Mass of calcium oxide = y g
Mass of magnesium oxide = (0.96 - y)g

Applying POAC for Ca atoms,

Moles of Ca atoms in $C a C O_{3}$ = moles of Ca atoms in CaO

$1 \times moles of C a C O_{3} = 1 \times$ moles of CaO
Molar masses of calcium carbonate and calcium oxide are 100 and 56, respectively
$\frac{x}{100} = \frac{y}{56}$

Again applying POAC for Mg atoms,

Moles of Mg in $M g C O_{3}$ = moles of Mg in MgO
$1 \times Moles of M g C O_{3} = 1 \times$ moles of MgO

The molar mass of magnesium carbonate and magnesium oxide are 84 and 40 respectively.
$\frac{1.84 - x}{84} = \frac{0.96 - y}{40}$

From eqn (i) and (ii),
we get x = 0.83 g, y = 0.465 g
$% o f C a C O_{3} = \frac{0.83}{1.84} \times 100 = 45.11 % % o f M g C O_{3} = 54.89 %$

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1.84 g of a mixture of CaCO3 and MgCO3 was heated to a constant weight. The constant weight of the residue was found to be 0.96 g. calculate the percentage composition of the mixture.. (Ca=40,Mg=24,C=12,O=16)

1.84 g of a mixture of $C a C O_{3}$ and $M g C O_{3}$ was heated to a constant weight. The constant weight of the residue was found to be 0.96 g. calculate the percentage composition of the mixture.. $(C a = 40, M g = 24, C = 12, O = 16)$