Calculate the percentage composition of the elements present in magnesium carbonate. How many Kg of `CO_(2)` can be obtained from 100 Kg of is 90% pure magnesium carbonate

Explanation

Molar mass of MgCO3 = 84.32 g `mol^(-1)`
Percentage of Mg = `24/84.32 xx 100 = 28.46%`
Percentage of C =`12/84.32 xx 100` = 14.23%
Percentage of `O_(3) =48/84.32 xx 100 = 57.0%`
`MgCO_(3) rarr MgO + CO_(2)`
84.32 g of 100% pure `MgCO_(3)` gives 44g of `CO_(2)`
`:. 100xx10^(3)` g of 100 % pure `MgCO_(3)` gives =`44/84.32 xx 100 xx 10^(3)`
= `52.182 xx 10^(3)` g`CO_(2)`
100% pure` MgCO_(3)` gives `52.182 xx 10^(3)gCO_(2)`
`:.` 90% pure `MgCO_(3)` will give`(52.182xx10^(3))/100 xx 90` = 46963.8 g `CO_(2)`
= 46.96 Kg `CO_(2)`.