# Question 1f048

Mar 10, 2016

86.98%

#### Explanation:

In order to find the percent yield of a given reaction, you need to know two things

• the theoretical yield of the reaction, i.e .how much product you would get if the reaction had a 100% yield
• the actual yield of the reaction, i.e. how much product you actually get

In your case, you know that the reaction produced $\text{18.25 g}$ of iron. This will be your actual yield.

In order to find the theoretical yield, you need to use the mole ratio that exists between iron(III) oxide, ${\text{Fe"_2"O}}_{3}$, and iron metal, $\text{Fe}$.

The balanced chemical equation for this reaction looks like this

${\text{Fe"_2"O"_text(3(s]) + 2"Al"_text((s]) -> "Al"_2"O"_text(3(s]) + color(red)(2)"Fe}}_{\textrm{\left(s\right]}}$

Notice that every mole of iron(III) oxide that takes part in the reaction will produce $\textcolor{red}{2}$ moles of iron metal.

Since aluminium is said to be in excess, you can assume that all the moles of iron(III) oxide will take part in the reaction. Use the compound's molar mass to determine how many moles you have in that $\text{30.00-g}$ sample of iron(III) oxide

30.00color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_3)/(159.69color(red)(cancel(color(black)("g")))) = "0.18786 moles Fe"_2"O"_3

So, if this is how many moles of iron(III) oxide take part in the reaction, you can use the aforementioned $1 : \textcolor{red}{2}$ mole ratio to figure out how many moles of iron would theoretically be produced by the reaction.

0.18786color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * overbrace((color(red)(2)color(white)(a)"moles Fe")/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))))^(color(purple)("1:2 mole ratio")) = "0.37572 moles Fe"

This many moles would correspond to a mass of

0.37572color(red)(cancel(color(black)("moles Fe"))) * overbrace("55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))))^(color(brown)("molar mass of Fe")) = "20.982 g Fe"

So, the reaction should theoretically produce $\text{20.982 g}$ of iron metal, but it ends up producing $\text{18.25 g}$. The percent yield of the reaction can be calculated using the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% yield" = "actual yield"/"theoretical yield} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to get

"% yield" = (18.25 color(red)(cancel(color(black)("g"))))/(20.982color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"86.98%"color(white)(a/a)|)))#

The answer is rounded to four sig figs.