1.0 g of a mixture of carbonates of calcium and magnesium gave 240 cm^(3) of CO_(2) at STP. 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 STP. Calculate the percentage composition of the mixture.
Answer» Solution :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` .