( 1.0 ~g ) of ( Mg ) is burnt with ( 0.28 ~g ) of ( O ₂ ) in a closed vessel. Which reactant is left in excess and how much?

Step-by-step solution

For the reaction 2 Mg + O ₂ 2 M gO 48 ~g of Mg requires 32 ~g of O ₂ Therefore, 1 ~g of Mg require 32 48 ~g of O ₂=0.66 ~g But available O ₂ is 0.28 ~g So, O ₂ is the limiting reagent and Mg is the excess reagent. 32 ~g of O ₂ needs 48 ~g of Mg 1 ~g of O ₂ needs 48 32 ~g of Mg 0.28 of O ₂ needs 48 32 0.28=0.42 ~g Mg available =1 ~g Mg left =1-0.42=0.58 ~g