Question

A sample containing only $\,\,CaC{O_3}\,$ and $\,\,MgC{O_3}\,$ is ignited to $\,\,CaO\,\,$ and $\,\,MgO\,\,$. The mixtures of oxides produced weight exactly half as much as the original sample. The mass percentages of $\,\,CaC{O_3}\,$ and $\,MgC{O_3}\,$ in the sample are respectively:
A.$\,\,28.4\% ,71.6\% \,\,$
B.$\,\,71.6\% ,28.4\% \,\,$
C.$\,\,56.7\% ,43.3\% \,\,$
D.$\,\,78.5\% ,21.5\% \,\,$

Answer 1

Hint:Mass per cent is the way a concentration is represented. In addition, in a specific mixture, it defines the element. For the mass of any element in one mole of the compound, the mass percent formula is expressed to find out the molar mass.
Here, the total mass of the sample is considered as $\,\,100g\,\,$.
Formula used:
There is only one formula used in this entire question, that is
$\,\,n = \dfrac{m}{{GMM}}\,\,$ Here, $\,\,n\,\,$ stands for number of moles
$\,\,m\,\,$ stands for given mass
And $\,\,GMM\,\,$ stands for gram molecular mass

Complete step by step answer:
Mass of $\,\,Ca = 40\,\,$,$\,\,C = 12,O = 16,Mg = 24.3\,\,$
 $\,\,GMM\,\,$ of the respective compounds;
$\,\,CaC{O_3}\,$$\,\, = 40 + 12 + 3 \times 16 = 40 + 12 + 48 = 100\,\,$
 $\,\,MgC{O_3} = 24.3 + 12 + 3 \times 16 = 24.3 + 12 + 48 = 84.3\,\,$
  $\,\,CaO = 40 + 16 = 56\,\,$
  $\,\,MgO = 24.3 + 16 = 40.3\,\,$
Now, let the total mass of sample be $\,\,100g\,\,$. Let mass of $\,\,CaC{O_3}\,$ be $\,\,x\,\,$ and that of $\,\,MgC{O_3}\,$ is $\,\,100 - x\,\,$.
$\,\,CaC{O_3} \to CaO + C{O_2}\,\,$ [ equation $\,1\,$]
Number of moles of $\,\,CaC{O_3}\,$$\,\,\, = \dfrac{x}{{100}}\,\,\,$, so number of moles of $\,\,\,CaO\,\,\,$ is also $\,\,\,\dfrac{x}{{100}}\,\,\,$.
So, mass of $\,\,\,CaO\,\,\,$$\, = \,\dfrac{x}{{100}}\, \times 56\,$ [ because $\,\,n = \dfrac{m}{{GMM}}\,\,$, so $\,\,n \times GMM = m\,\,$]
$\,\,MgC{O_3} \to MgO + C{O_2}\,\,\,$[ equation $\,2\,$]
Number of moles of $\,\,MgC{O_3}\,$ is $\,\,\dfrac{{100 - x}}{{84.3}}\,\,$, so number of moles of $\,\,MgO\,\,$is also $\,\,\dfrac{{100 - x}}{{84.3}}\,\,$.
So, mass of $\,\,MgO\,\,$$\, = \,\dfrac{{100 - x}}{{84.3}}\, \times 40.3\,$
So, now according to the given condition that says that the mass of $\,\,CaO\,\,$ and $\,\,MgO\,\,$ is going to be half of the total mass of the sample that is $\,\,\dfrac{{100}}{2} = 50\,\,$.
So, therefore $\,\,\dfrac{{100 - x}}{{84.3}}\, \times 40.3 + \dfrac{x}{{100}} \times 56 = 50\,$
Calculating this will give $\,\,x = 28.4\,\,$ that is mass percentage of $\,\,CaC{O_3}\,$ $\,\, = 28.4\% \,\,$ and $\,\,100 - x = 71.6\,\,$ that is mass percentage of $\,\,MgC{O_3}\,$ $\,\, = 71.6\% \,\,$
So, the correct answer is Option A. Hope this explanation cleared all your doubts.

Note:At standard temperature the molar volume is the volume occupied by $\,\,1\,mole\,$ of a chemical element or compound. It is calculated by dividing the molar mass by mass density. Molar volume of a gas at standard temperature and pressure, is equal to $\,\,22.4\,litres\,$ for \[\,\,1mole\,\] of any ideal gas at temperature $\,\,273.15K\,\,$ and $\,\,1atm\,\,$ pressure.