Question

11.775 g of magnesium is reacted with 4.938 g of oxygen gas according to the equation: 2 Mg + O2 --> 2 MgO, what is the limiting reactant?

Answer 1

The limiting reagent is oxygen gas.

`2Mg(s)+O_2(g)rarr2MgO(s)`

Explanation:

Moles of `O_2` `=` `(4.938*g)/(32.00*g*mol^-1)` `=` `0.154` `mol`.

Moles of `Mg` `=` `(11.775*g)/(24.305*g*mol^-1)` `=` `0.484` `mol`.

By the reaction stoichiometry, magnesium is clearly the reagent in excess. We would require 0.968 mol of oxygen gas for stoichiometric reaction