Calculate the percentage composition of the elements present in magnesium carbonate. How many kilogram of `CO_(2)` can be obtained by heating 1 kg of 90 % pure magnesium carbonate.

Explanation

The balanced chemical equation is
`MgCO_(3)overset(Delta)to MgO +CO_(2)`
Molar mass of `MgCO_(3)` is 84 g `"mol"^(-1)`.
84 g `MgCO_(3)` contain 24 g of Magnesium.
`:.` 100 g of `MgCO_(3)` contain
`(24g Mg)/(cancel(84g MgCO_(3))) xx cancel(100g MgCO_(3))`
`=28.57g Mg`.
i.e. percentage of magnesium
`=28.57%`
`84 g MgCO_(3)` contain 12 g of carbon

`:. 100 g MgCO_(3)` contain

`=14.29` g of carbon.
`:.` Percentage of carbon
`=14.29%`
`84g MgCO_(3)` contain 48 g of oxygen
`:. 100 g MgCO_(3)` contains

`=57.14` g of oxygen.
`:.` Percentage of oxygen
`=57.14%`.
As per the stoichiometric equation,
84 g of `100%` pure `MgCO_(3)` on heating gives 44g of `CO_(2)`.
`:.` 1000 g of 90% pure `MgCO_(3)` gives

`=471.43g " of "CO_(2)`
= 0.471 kg of `CO_(2)`