Question

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 1

Correct Answer - `I_(2)`
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.