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

`CaCO_(3)=62.5 %, MgCO_(3)=37.5 %`

Explanation

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 %`.