Question

$92g$ mixture of $CaC{O_3}$ and $MgC{O_3}$ heated strongly in an open vessel. After complete decomposition of the carbonates it was found that the weight of residue left behind is $48g$ . Find the mass of $MgC{O_3}$ in grams in the mixture.

Answer 1

Calcium carbonate is a white colored powdered compound which on strong heating (calcination) produces calcium oxide along with a release of carbon dioxide. Magnesium carbonate is also a colorless powdered compound which on thermal decomposition produces magnesium oxide and releases carbon dioxide gas.

Complete answer:
According to the question, a mixture of calcium carbonate and magnesium carbonate is heated strongly. The reaction that takes place is as follows:
$CaC{O_3}(s) + MgC{O_3}(s)\xrightarrow{\Delta }CaO(s) + MgO(s) + 2C{O_2}(g)$
From the above reaction, we can see that the thermal decomposition of calcium carbonate releases calcium oxide and carbon dioxide and the thermal decomposition of magnesium carbonate releases magnesium oxide and carbon dioxide.
Thus, one mole of $CaC{O_3}$ produces = one mole of $CaO$
And, one mole of $MgC{O_3}$ produces = one mole of $MgO$
The molecular weight of $CaC{O_3} = 40 + 12 + (16 \times 3) = 100g$
The molecular weight of $MgC{O_3} = 24 + 12 + 48 = 84g$
The molecular weight of $CaO = 40 + 16 = 56g$
The molecular weight of $MgO = 24 + 16 = 40g$
When a mixture of $184g$ is heated, it produces = $(56 + 40)g$ of product
When a mixture of $1g$ is heated, it produces = $\dfrac{{96}}{{184}}g$ of product
When a mixture of $92g$ is heated, it produces = $\dfrac{{96 \times 92}}{{184}} = 48g$ of product
Let the weight of $MgC{O_3}$ in the mixture = $x$
Then, the weight of $CaC{O_3}$ in the mixture = $(92 - x)g$
The residue given by $84{\text{ }}gm$ of $MgC{O_3}$ $= {\text{ }}40{\text{ }}gm$
The residue given by $x$ grams of $MgC{O_3}$ $= {\text{ }}\left( {\dfrac{{40}}{{84}}} \right) \times x{\text{ }}gm$
The residue given by $100{\text{ }}gm$ of $CaC{O_3}$ = $56{\text{ }}gm$
The residue given by $\left( {92{\text{ }} - {\text{ }}x} \right){\text{ }}gm$ of $CaC{O_3}$ $= \left( {\dfrac{{56}}{{100}}} \right) \times \left( {92 - x} \right){\text{ }}gm$
According to the question, we have:
The weight of residue = $48{\text{ }}gm$
Thus, the equation for the residue left in the reaction is as follows:
\Rightarrow \left\\{ {\left( {\dfrac{{40}}{{84}}} \right) \times x} \right\\} + \left\\{ {\left( {\dfrac{{56}}{{100}}} \right) \times \left( {92 - x} \right)} \right\\} = 48
$\Rightarrow 0.476{\text{ }}x + 51.52 - 0.56{\text{ }}x = 48$ ⇒ 0.084 x = 3.52
$\Rightarrow x = \dfrac{{3.52}}{{0.084}} = 41.9gm$
So, the mass of $MgC{O_3}$ in the mixture is $41.9{\text{ }}gm$ .

Note:
POAC stands for Principle of Atomic Conservation which is used to solve a chemical equation. There is no need of balancing the equation if we adopt POAC. It is based on the law of conservation of mass of an atom or a mole of an atom. The stoichiometry for the above reaction can also be determined from the POAC.