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1.84 g of a mixture of CaCO_(3) and MgCO_(3) is strongly heated till no further loss in mass takes place. The residue is found to weigh 0.96 g. Calculate the percentage of each component in the mixture.

Answer» <html><body><p></p>Solution :Both `CaCO_(3) and MgCO_(3)` upon <a href="https://interviewquestions.tuteehub.com/tag/heating-1017297" style="font-weight:bold;" target="_blank" title="Click to know more about HEATING">HEATING</a> <a href="https://interviewquestions.tuteehub.com/tag/lose-537625" style="font-weight:bold;" target="_blank" title="Click to know more about LOSE">LOSE</a> `CO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` and the mass of residue is that of CaO and MgO. Let us calculate the same from the mixture and quate with the given mass `(0.96g)` <br/> Let the mass of `CaCO_(3)` in the mixture =xg <br/> `:.` Mass of `MgCO_(3)` in the mixture `(1.84 - x)g` <br/> Step I. Calculation of mass of residue (CaO) from x g of `CaCO_(3)` <br/> The chemical equation for the <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> is : <br/> `{:(CaCO_(3),overset("Heat")(rarr),CaO+CO_(2)),(40+12+48,,40+16),(=100,,=56g):}`<br/> 100 g of `CaCO_(3)` upon heating from CaO (residue) = 56 g <br/> `:.` x g of `CaCO_(3)` upon heating form CaO (residue) `= (56)/(100) xx xg` <br/> Step II. Calculation of mass of residue (MgO) from `(1-x)`of `MgCO_(3)` <br/> `{:(MgCO_(3),overset("Heat")(rarr),MgO+CO_(2)),(24+12+48,,24+16),(=84g,,=40g):}` <br/> 84 g of `MgCO_(3)` upon heating from MgO (residue) = 40 g <br/> `(1-x)` of `MgCO_(3)` upon heating from MgO (residue) `= (40)/(84) xx(1-x)g` <br/> Step III. Calculation of percentage composition of the mixture <br/> Total mass of the residue `= [(56)/(100)xxx+(40)/(84)(1-x)]g` <br/> But the mass of residue actually formed = 0.96 g <br/> <a href="https://interviewquestions.tuteehub.com/tag/equating-7679226" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATING">EQUATING</a> the two, we get <br/> `(56x)/(100)+(40(1-x))/(84)=0.96` <br/> `4704x+7360-4000x=0.96xx100xx84=8064` <br/> `704x=8064-7360=704,x=(704)/(704)=1` <br/> Mass of `CaCO_(3)` in the mixture = 1g <br/> Mass of `MgCO_(3)` in the mixture `= 1.84 - 1 = 0.84`g <br/> Percentage of `CaCO_(3)` in the mixture `= (1)/(1.84) xx 100 = 54.35 %` <br/> Percentage of `MgCO_(3)` in the mixture `= (0.84)/(1.84) xx 100 = 45.65 %`</body></html>


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