# Question 9b2b7

Oct 31, 2015

Clearly ferric oxide is present in deficiency, and this is the limiting reagent.

#### Explanation:

$3 M g \left(s\right) + F {e}_{2} {O}_{3} \left(s\right) \rightarrow 2 F e \left(s\right) + 3 M g O$

The stoichiometry of the reaction demands that 3 mol magnesium be present per 1 mol ferric oxide. Starting conditions are 1 mol magnesium, and only 0.252 mol ferric oxide. Clearly, magnesium is present in excess, and ferric oxide is the limiting reagent.

So how much magnesium oxide, and how much iron are you getting to isolate from the reaction?

Oct 31, 2015

The limiting reactant is $\text{Fe"_2"O"_3}$.
The theoretical yield is $\text{30.470 g MgO}$.

#### Explanation:

Balanced Equation

$\text{3Mg"+"Fe"_2"O"_3}$$\rightarrow$$\text{3MgO"+"2Fe}$

Magnesium as Reactant

Multiply the given $\text{1 mol Mg}$ times the mole ratio of $\text{MgO}$ to $\text{Mg}$ from the balanced equation : $\left(3 \text{mol MgO")/(3"mol Mg}\right)$.

$1 \cancel{\text{mol Mg"xx(cancel(3"mol MgO"))/(cancel(3"mol Mg"))="1 mol MgO}}$

This cancels moles of $\text{Mg}$, leaving moles of $\text{MgO}$.

Next multiply the moles $\text{MgO}$ times its molar mass, $\text{40.3044 g/mol}$. This can also be written as $\left(\text{40.3044 g MgO")/(1"mol MgO}\right)$.

$1 \cancel{\text{mol MgO"xx(40.3044"g MgO")/(1cancel"mol MgO")="40.3044 g MgO}}$

The maximum mass of $\text{MgO}$ that can be produced by $\text{1 mol Mg}$ in this reaction is $\text{40.3044 g}$.

Iron(III) Oxide as Reactant

Multiply the given ${\text{0.252 mol Fe"_2"O}}_{3}$ times the mole ratio of $\text{MgO}$ to $\text{Fe"_2"O"_3}$ from the balanced equation"$\left(3 \text{mol MgO")/(1"mol Fe"_2"O"_3}\right)$

$0.252 \cancel{\text{mol Fe"_2"O"_3xx(3cancel"mol MgO")/(1cancel"mol Fe"_2"O"_3)="0.756 mol MgO}}$

This cancels moles of $\text{Fe"_2"O"_3}$, leaving moles of $\text{MgO}$.

Multiply the $\text{0.756 mol MgO}$ times its molar mass of $\text{40.3044 g/mol}$.

$0.756 \cancel{\text{mol MgO"xx(40.3044"g MgO")/(1cancel"mol MgO")="30.470 g MgO}}$

The maximum mass of $\text{MgO}$ that can be produced by $\text{0.252 mol Fe"_2"O"_3}$ is $\text{30.470 g}$

Limiting Reagent

The maximum yield of $\text{MgO}$ from $\text{1 mol Mg}$ is 40.3044 g"#
The maximum yield of $\text{MgO}$ from $\text{0.252 mol Fe"_2"O"_3}$ is $\text{30.470 g}$

Therefore, the limiting reagent is $\text{Fe"_2"O"_3}$.

There is an excellent tutorial at the following website
https://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm