Question

Calculate the percentage composition of the elements present in magnesium carbonate. How many Kg of CO₂ can be obtained from 100 Kg of is 90% pure magnesium carbonate.

Answer 1

Molar mass of MgCO₃ = 84.32 g mol^-1

Percentage of Mg = \(\frac{24}{84.32}\) x 100 = 28.46% 

Percentage of C = \(\frac{12}{84.32}\) x 100 = 14.23% 

Percentage of O₃ = \(\frac{48}{84.32}\) x 100 = 57.0%

84.32 g of 100% pure MgCO₃ gives 44g of CO₂ 

∴ 100 x 10³ g of 100% pure MgCO₃ gives = \(\frac{44}{84.32}\) x 100 x 10³ 

= 52.182 x 10³ g CO₂ 

100% pure MgCO₃ gives 52.182 x 10³ g CO₂ 

∴ 90% pure MgCO₃ will give \(\frac{52.182\times10^3}{100}\) x 90 = 46963.8 g CO₂