Assume that 4g of I_(2) are allowed to react with 4g of Mg metal according to the following equation Mg+I_(2) rarr MgI_(2) Which is the limiting reagent in the reaction ?

Assume that 4g of I_(2) are allowed to react with 4g of Mg metal accor

1.	Assume that 4g of I_(2) are allowed to react with 4g of Mg metal according to the following equation Mg+I_(2) rarr MgI_(2) Which is the limiting reagent in the reaction ?
Answer» Solution :The chemical EQUATION is : `underset(24g)(Mg)+underset(underset(=254g)(2xx127))(I_(2))rarrMgl_(2)` 24 G of Mg metal require `I_(2) = 254 g` 4 g of Mg metal require `I_(2)=((254g))/((24g))XX(4g)=42.3g` But `I_(2)` actually available = 4g `:. I_(2)` is the limiting reactant.