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)rarr` 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
= ` (24 g Mg)/(84 g MgCO_(3)) xx 100 g MgCO_(3) `
= 28.57 g Mg i.e. percentage of magnesium = 28.57 % .
84 g `MgCO_(3)` contain 12 g of carbon

`:. ` 100g `MgCO_(3)` contain
`(12 g C)/(84 g MgCO_(3)) xx 100 g MgCO_(3)`
= 14.29 g of carbon
`:. ` percentage of carbon =14.29%
84 g `MgCO_(3)` contains 48 g of oxygen
`:.` 100 g `MgCO_(3) `contains
`(48 g O)/(84 g MgCO_(3)) xx 100g MgCO_(3)`
= 57.14 g of oxygen
`:.` percentage of oxygen = 57.14%
As per the stoichiometric equation,
84 g 100% pure `MgCO_(3)` on heating gives 44 g of `CO_(2)`
`:. ` 1000 g of 90% pure `MgCO_(3)` gives
`(44 g)/(84 g xx 100%)xx 90% xx 1000`g
= 471.43 g of `CO_(2)`
= 0.471 kg of `CO_(2)`