# 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?

Dec 5, 2015

The limiting reagent is oxygen gas.

$2 M g \left(s\right) + {O}_{2} \left(g\right) \rightarrow 2 M g O \left(s\right)$

#### Explanation:

Moles of ${O}_{2}$ $=$ $\frac{4.938 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.154$ $m o l$.

Moles of $M g$ $=$ $\frac{11.775 \cdot g}{24.305 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.484$ $m o l$.

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