# What is the limiting reagent and maximum yield if "0.404 mol Fe" reacts with "0.135 mol Cl"_2"?

## $\text{2Fe(s) + 3Cl"_2("g")}$$\rightarrow$$\text{2FeCl"_3}$

Jul 2, 2015

#### Answer:

The limiting reagent is $C {l}_{2}$.
The maximum amount of iron (III) chloride produced by this reaction is 0.0900 moles.

#### Explanation:

The balanced chemical equation for this reaction looks like this

$\textcolor{red}{2} F {e}_{\left(s\right)} + \textcolor{b l u e}{3} C {l}_{2 \left(g\right)} \to 2 F e C {l}_{3 \left(s\right)}$

Notice that you have a $\textcolor{red}{2} : \textcolor{b l u e}{3}$ mole ratio between iron and chlorine gas. This means that, regardless of how many moles of iron react, you'll always need 3/2 times more moles of chlorine gas in order for the reaction to take place.

If all the iron is to react, then the reaction would need

0.404cancel("moles"Fe) * (color(blue)(3)" moles "Cl_2)/(color(red)(2)cancel("moles"Fe)) = "0.606 moles" $C {l}_{2}$

Since you don't have enough chlorine gas to allow all the iron to react, chlorine gas, $C {l}_{2}$, will be the limiting reagent.

This means that the amount of chlorine you have will determine how much iron will actually take part in the reaction.

0.135cancel("moles"Cl_2) * (color(red)(2)" moles "Fe)/(color(blue)(3)cancel("moles"Cl_2)) = "0.0900 moles" $F e$

Notice that you have a $\textcolor{red}{1} : 1$ mole ratio between iron and iron (III) chloride. This means that the number of moles of iron (III) chloride produced by this reaction will be equal to the number of moles of iron that reacted.

0.0900cancel("moles"Fe) * ("2 moles "FeCl_3)/(color(red)(2)cancel("moles"Fe)) = "0.0900 moles" $F e C {l}_{3}$

If you want to see how many grams of iron (III) chloride you'd get, use the compound's molar mass

0.0900cancel("moles") * "162.204 g"/(1cancel("mole")) = "14.6 g" $F e C {l}_{3}$

Jul 2, 2015

#### Answer:

The limiting reagent is the chlorine gas, with the formula ${\text{Cl}}_{2}$.
The maximum amount of iron (III) chloride $\left(\text{FeCl"_3}\right)$ that can be formed is $\text{0.0900 mol}$ or $\text{14.59 g}$.

#### Explanation:

Balanced Equation

"2Fe(s)"+"3Cl"_2("g")$\rightarrow$"2FeCl"_3("s")

In words: Two moles of solid iron plus three moles of chlorine gas produce two moles ofsolid iron (III) chloride.

Determine the mole ratios of each substance from the balanced equation.

Mole ratio of $\text{Fe}$ and $\text{FeCl"_3}$.

$\left(\text{2 mol Fe")/("2 mol FeCl"_3}\right)$ and $\left(\text{2 mol FeCl"_3)/("2 mol Fe}\right)$

Mole ratio of ${\text{Cl}}_{2}$ and $\text{FeCl"_3}$

$\left({\text{3 mol Cl"_3)/("2 mol FeCl}}_{3}\right)$ and $\left(\text{2 mol FeCl"_3)/("3 mol Cl"_3}\right)$

Moles of ${\text{FeCl}}_{3}$ produced by $\text{Fe}$.

Determine the number of moles of ${\text{FeCl}}_{3}$ produced by $\text{0.404 mol Fe}$ by multiplying the given moles of $\text{Fe}$ times the mole ratio with ${\text{FeCl}}_{3}$ in the numerator.

$0.404 {\cancel{\text{mol Fe"xx("2 mol FeCl"_3)/(2 cancel"mol Fe")="0.404 mol FeCl}}}_{3}$

Moles of ${\text{FeCl}}_{3}$ produced by $\text{Cl"_2}$.

Determine the number of moles of ${\text{FeCl}}_{3}$ produced by $\text{0.135 mol Cl"_3}$ by multiplying the given moles of $\text{Cl"_2}$ times the mole ratio with ${\text{FeCl}}_{3}$ in the numerator.

$0.135 \cancel{\text{mol Cl"_2xx("2 mol FeCl"_3)/(3 cancel"mol Cl"_2)="0.0900 mol FeCl"_3}}$

Determine the limiting reagent.

The ${\text{Cl}}_{2}$ is the limiting reagent because it produces fewer moles of $\text{FeCl"_3}$. To determine the maximum amount in grams of ${\text{FeCl}}_{3}$ that can be produced, multiply the moles of ${\text{FeCl}}_{3}$ possible times its molar mass, which is $\text{162.204 g/mol}$.

$0.0900 \cancel{\text{mol FeCl"_3xx("162.204 g FeCl"_3)/(1 cancel"mol FeCl"_3)="14.59g FeCl"_3}}$

Complete Answer

The limiting reagent is $\text{Cl"_2}$, which will produce ${\text{0.0900 mol FeCl}}_{3}$ from ${\text{0.135 mol Cl}}_{2}$.

The maximum amount in grams of $\text{FeCl"_3}$ that can be produced from ${\text{0.0900 mol Cl}}_{2}$ is $\text{14.59 g FeCl"_3}$.