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.

1.0 g of a mixture of carbonates of calcium and magnesium gave 240 cm^

1.	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» Solution :Let the mass of `CaCO_(3)` in the MIXTURE =x g `:.` The mass of `MgCO_(3)` in the mixture `= (1-x)g` Step I. Volume of `CO_(2)` evolved from `CaCO_(3)` `underset(40+12+48+100g)(CaCO_(3)(s))overset("Heat")(rarr)CAO(s)+underset(22400"cc")(CO_(2)(g))` 100 g of `CaCO_(3)` evolve `CO_(2)` upon heating = 22400 cc xg of `CaCO_(3)` evolve `CO_(2)` upon upon heating `= (22400)/(100) xx x c c = (224x)c c`. Step II. Volume of `CO_(2)` evolved from `MgCO_(3)` `underset(24+12=48=84g)(MgCO_(3)(s))overset("Heat")(rarr)MgO(s)+underset("22400cc")(CO_(2)(g))` 84 g of `MgCO_(3)` evolve `CO_(2)` upon heating = 22400 cc `(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` Step III. Percentage composition of the mixture According to available data : `224 x + 266.6 (1-x)=240` `224 x + 266.6 - 266.6 x = 240` `- 42.6 x= -26.6 or x = (26.6)/(42.6)=0.625` `:.` Percentage of `CaCO_(3)` in the mixture `= 0.625 xx 100 = 62.5 %` Percentage of `MgCO_(3)` in the mixture `= 100 - 62.5 = 37.5 %`.