The reaction between magnesium (Mg) and oxygen (O₂) to form magnesium oxide (MgO) can be represented by the following balanced chemical equation:
\[ 2 Mg + O_2 \rightarrow 2 MgO \]
This equation tells us that two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide. We can use this stoichiometric relationship to determine the limiting reactant when given the amounts of reactants available.
We have:
- 4.0 moles of magnesium
- 3.0 moles of oxygen
From the stoichiometry of the reaction, we can see that the mole ratio of magnesium to oxygen is 2:1. For each mole of oxygen, we need two moles of magnesium to completely react.
Now let's calculate the moles of magnesium needed to react with all of the oxygen:
3.0 moles of oxygen (O₂) x (2 moles of Mg / 1 mole of O₂) = 6.0 moles of Mg
This calculation tells us that to fully react with 3.0 moles of oxygen, we would need 6.0 moles of magnesium. However, we only have 4.0 moles of magnesium available, which is not enough to react with all the oxygen present.
Since magnesium is the reactant we would run out of first (as we require 6.0 moles but only have 4.0 moles), magnesium is the limiting reactant.
Now we'll calculate the amount of magnesium oxide formed by the reaction of 4.0 moles of magnesium with oxygen:
Because magnesium is the limiting reactant, we base our calculation of the product formed on the amount of magnesium available. From the balanced equation, we see that 2 moles of magnesium will produce 2 moles of magnesium oxide. Therefore, 4.0 moles of magnesium will produce:
4.0 moles of Mg x (2 moles of MgO / 2 moles of Mg) = 4.0 moles of MgO
So, the limiting reactant in this reaction is magnesium (Mg), and the amount of product (magnesium oxide, MgO) formed is 4.0 moles.
