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1.0 g of a mixture of carbonates of calcium and magnesium gave 240 cm^(3) of CO_(2) at N.T.P Calculate the percentage composition of the mixture.

Answer» <html><body><p><br/></p>Solution :Let the mass of `CaCO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` in the <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> =x g <br/> `:.` The mass of `MgCO_(3)` in the mixture `= (1-x)g` <br/> Step I. Volume of `CO_(2)` evolved from `CaCO_(3)` <br/> `underset(40+12+48+100g)(CaCO_(3)(s))overset("Heat")(rarr)<a href="https://interviewquestions.tuteehub.com/tag/cao-411869" style="font-weight:bold;" target="_blank" title="Click to know more about CAO">CAO</a>(s)+underset(22400"cc")(CO_(2)(g))` <br/> 100 g of `CaCO_(3)` evolve `CO_(2)` upon heating = 22400 cc <br/> xg of `CaCO_(3)` evolve `CO_(2)` upon upon heating `= (22400)/(100) xx x c c = (224x)c c`. <br/> Step II. Volume of `CO_(2)` evolved from `MgCO_(3)` <br/> `underset(24+12=48=84g)(MgCO_(3)(s))overset("Heat")(rarr)MgO(s)+underset("22400cc")(CO_(2)(g))` <br/> 84 g of `MgCO_(3)` evolve `CO_(2)` upon heating = 22400 cc <br/> `(1-x)`g of `MgCO_(3)` evolve `CO_(2)` upon heating `= 22400 xx ((1-x))/(84) c c. = 266.6 xx (1-x)c c` <br/> Step III. Percentage composition of the mixture <br/> According to available data : <br/> `224 x + 266.6 (1-x)=240` <br/> `224 x + 266.6 - 266.6 x = 240` <br/> `- 42.6 x= -26.6 or x = (26.6)/(42.6)=0.625` <br/> `:.` Percentage of `CaCO_(3)` in the mixture `= 0.625 xx 100 = 62.5 %` <br/> Percentage of `MgCO_(3)` in the mixture `= 100 - 62.5 = 37.5 %`.</body></html>


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