1 gm of a mixture of calcium carbonate and magnesium carbonate gave on ignition 240 mL of carbon dioxide at S.T.P. What is the percentage composition of the mixture ?

Explanation

Mass of mixture of carbonates of Ca and Mg taken = 1.0 g
Suppose the mass of `CaCO_(3) = x g " " therefore` Mass of `MgCO_(3)=(1-x)g`
The chemical equations involved are : `underset(40+12+3xx16=100g)(CaCO_(3))rarr CaO+underset(22400 cm^(3) "at STP")(CO_(2))` ....(i)
`underset(24+12+3xx16=84 g)(MgCO_(3))rarr MgO+underset(22400 cm^(3) "at STP")(CO_(2))` ....(ii)
Step 1. To calculate the volume of `CO_(2)` evolved at STP from x g of `CaCO_(3)`.
100 g of `CaCO_(3)` evolve `CO_()` at STP `= 22400 cm^(3)`
`therefore x` g of `CaCO_(3)` will evolve `CO_(2)` at STP `= (22400)/(100)xx x cm^(3)`
Step 2. To calculate the volume of `CO_(2)` evolved at STP (1 - x) g of `MgCO_(3)`.
84 g of `MgCO_(3)` evolve `CO_(2)` at STP `= 22400 cm^(3)`
`therefore (1-x)` g of `MgCO_(3)` will evolve `CO_(2)` at STP `=(22400)/(84)xx(1-x)cm^(3)=(800)/(3)(1-x)cm^(3)`
Step 3. To calculate the value of x
Total volume of `CO_(2)` evolved at STP `= 224 x + (800)/(3)(1-x)cm^(3)`
But total volume of `CO_(2)` evolved at STP `= 240 cm^(3)` (Given)
`therefore 224 x+(800)/(3)(1-x)=240` or `672 x + 800-800x = 720` or 128 x = 80
`therefore x= (5)/(8)`
Step 4. To calculate the percentage composition of the mixture.
`therefore` Percentage of `CaCO_(3)=(5)/(8xx1)xx100=62.5`
`therefore` Percentage of `MgCO_(3)=100-62.5=37.5` .