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Question: A sample containing only \(\,\,CaC{O_3}\,\) and \(\,\,MgC{O_3}\,\) is ignited to \(\,\,CaO\,\,\) and...

A sample containing only   CaCO3 \,\,CaC{O_3}\, and   MgCO3 \,\,MgC{O_3}\, is ignited to   CaO  \,\,CaO\,\, and   MgO  \,\,MgO\,\,. The mixtures of oxides produced weight exactly half as much as the original sample. The mass percentages of   CaCO3 \,\,CaC{O_3}\, and  MgCO3 \,MgC{O_3}\, in the sample are respectively:
A.  28.4%,71.6%  \,\,28.4\% ,71.6\% \,\,
B.  71.6%,28.4%  \,\,71.6\% ,28.4\% \,\,
C.  56.7%,43.3%  \,\,56.7\% ,43.3\% \,\,
D.  78.5%,21.5%  \,\,78.5\% ,21.5\% \,\,

Explanation

Solution

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  \,\,100g\,\,.
Formula used:
There is only one formula used in this entire question, that is
  n=mGMM  \,\,n = \dfrac{m}{{GMM}}\,\, Here,   n  \,\,n\,\, stands for number of moles
  m  \,\,m\,\, stands for given mass
And   GMM  \,\,GMM\,\, stands for gram molecular mass

Complete step by step answer:
Mass of   Ca=40  \,\,Ca = 40\,\,,  C=12,O=16,Mg=24.3  \,\,C = 12,O = 16,Mg = 24.3\,\,
  GMM  \,\,GMM\,\, of the respective compounds;
  CaCO3 \,\,CaC{O_3}\,   =40+12+3×16=40+12+48=100  \,\, = 40 + 12 + 3 \times 16 = 40 + 12 + 48 = 100\,\,
  MgCO3=24.3+12+3×16=24.3+12+48=84.3  \,\,MgC{O_3} = 24.3 + 12 + 3 \times 16 = 24.3 + 12 + 48 = 84.3\,\,
  CaO=40+16=56  \,\,CaO = 40 + 16 = 56\,\,
  MgO=24.3+16=40.3  \,\,MgO = 24.3 + 16 = 40.3\,\,
Now, let the total mass of sample be   100g  \,\,100g\,\,. Let mass of   CaCO3 \,\,CaC{O_3}\, be   x  \,\,x\,\, and that of   MgCO3 \,\,MgC{O_3}\, is   100−x  \,\,100 - x\,\,.
  CaCO3→CaO+CO2  \,\,CaC{O_3} \to CaO + C{O_2}\,\, [ equation  1 \,1\,]
Number of moles of   CaCO3 \,\,CaC{O_3}\,    =x100   \,\,\, = \dfrac{x}{{100}}\,\,\,, so number of moles of    CaO   \,\,\,CaO\,\,\, is also    x100   \,\,\,\dfrac{x}{{100}}\,\,\,.
So, mass of    CaO   \,\,\,CaO\,\,\,  = x100 ×56 \, = \,\dfrac{x}{{100}}\, \times 56\, [ because   n=mGMM  \,\,n = \dfrac{m}{{GMM}}\,\,, so   n×GMM=m  \,\,n \times GMM = m\,\,]
  MgCO3→MgO+CO2   \,\,MgC{O_3} \to MgO + C{O_2}\,\,\,[ equation  2 \,2\,]
Number of moles of   MgCO3 \,\,MgC{O_3}\, is   100−x84.3  \,\,\dfrac{{100 - x}}{{84.3}}\,\,, so number of moles of   MgO  \,\,MgO\,\,is also   100−x84.3  \,\,\dfrac{{100 - x}}{{84.3}}\,\,.
So, mass of   MgO  \,\,MgO\,\,  = 100−x84.3 ×40.3 \, = \,\dfrac{{100 - x}}{{84.3}}\, \times 40.3\,
So, now according to the given condition that says that the mass of   CaO  \,\,CaO\,\, and   MgO  \,\,MgO\,\, is going to be half of the total mass of the sample that is   1002=50  \,\,\dfrac{{100}}{2} = 50\,\,.
So, therefore   100−x84.3 ×40.3+x100×56=50 \,\,\dfrac{{100 - x}}{{84.3}}\, \times 40.3 + \dfrac{x}{{100}} \times 56 = 50\,
Calculating this will give   x=28.4  \,\,x = 28.4\,\, that is mass percentage of   CaCO3 \,\,CaC{O_3}\,   =28.4%  \,\, = 28.4\% \,\, and   100−x=71.6  \,\,100 - x = 71.6\,\, that is mass percentage of   MgCO3 \,\,MgC{O_3}\,   =71.6%  \,\, = 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 \,\,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 \,\,22.4\,litres\, for   1mole \,\,1mole\, of any ideal gas at temperature   273.15K  \,\,273.15K\,\, and   1atm  \,\,1atm\,\, pressure.