Answer:
[tex]\boxed{\text{Mg is the limiting reactant}}[/tex]
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble the data in one place.
2Mg + O₂ ⟶ 2MgO
n/mol: 2 5
Calculate the moles of MgO we can obtain from each reactant.
From Mg:
The molar ratio of MgO:Mg is 2:2
[tex]\text{Moles of MgO} = \text{2 mol Mg} \times \dfrac{\text{2 mol MgO}}{\text{2 mol Mg}} = \text{2 mol MgO}[/tex]
From O₂:
The molar ratio of MgO:O₂ is 2:1.
[tex]\text{Moles of MgO} = \text{5 mol O}_{2} \times \dfrac{\text{2 mol MgO}}{\text{1 mol O}_{2}} = \text{10 mol MgO}\\\\\boxed{\textbf{Mg is the limiting reactant}} \text{ because it gives the smaller amount of MgO}[/tex]
