Question

question 2 (1 point)
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limitproduct3. this reaction is about the following reaction:
o₂ + 2 mg → 2 mgo
1.5 moles of oxygen and 2.0 moles of magnesium are mixed for the reaction. how many moles of magnesium oxide will be produced?
4.0 moles
3.0 moles
6.0 moles
2.0 moles
1.0 moles
5.0 moles

Explanation:

Step1: Determine the limiting reactant

From the balanced equation $\ce{O_{2} + 2Mg -> 2MgO}$, the mole ratio of $\ce{O_{2}}$ to $\ce{Mg}$ is $1:2$.
Given 1.5 moles of $\ce{O_{2}}$ and 2.0 moles of $\ce{Mg}$.
If all $\ce{Mg}$ (2.0 mol) reacts, moles of $\ce{O_{2}}$ required = $\frac{2.0}{2} = 1.0$ mol (which is less than 1.5 mol available).
If all $\ce{O_{2}}$ (1.5 mol) reacts, moles of $\ce{Mg}$ required = $1.5\times2 = 3.0$ mol (but only 2.0 mol is available).
Thus, $\ce{Mg}$ is the limiting reactant.

Step2: Calculate moles of $\ce{MgO}$ produced

From the equation, 2 moles of $\ce{Mg}$ produce 2 moles of $\ce{MgO}$ (mole ratio $2:2 = 1:1$).
Since 2.0 moles of $\ce{Mg}$ (limiting reactant) react, moles of $\ce{MgO}$ produced = 2.0 moles.

Answer:

2.0 moles (corresponding to the option "2.0 moles")