# If 20 g of iron reacted with oxygen to form 20 g of iron(III) oxide, what is the percent of iron reacted?

Sep 3, 2017

The percent of iron reacted is 70 %.

#### Explanation:

Step 1. Write the chemical equation

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m} 55.84 \textcolor{w h i t e}{m m m m l l} 159.69$
$\textcolor{w h i t e}{m m m m l l} {\text{2Fe" + "3O"_2 → "Fe"_2"O}}_{3}$
$\text{Mass/g:} \textcolor{w h i t e}{m} 20 \textcolor{w h i t e}{m m m m m m l l} 20$

Step 2. Calculate the moles of ${\text{Fe"_2"O}}_{3}$

${\text{Moles of Fe"_2"O"_3 = 20 color(red)(cancel(color(black)("g Fe"_2"O"_3))) × ("1 mol Fe"_2"O"_3)/(159.69 color(red)(cancel(color(black)("g Fe"_2"O"_3)))) = "0.125 mol Fe"_2"O}}_{3}$

Step 3. Calculate the moles of $\text{Fe}$

$\text{Moles of Fe" = 0.125 color(red)(cancel(color(black)("mol Fe"_2"O"_3))) × "2 mol Fe"/(1 color(red)(cancel(color(black)("mol Fe"_2"O"_3)))) = "0.250 mol Fe}$

Step 4. Calculate the mass of $\text{Fe}$

$\text{Mass of Fe" = 0.250 color(red)(cancel(color(black)("mol Fe"))) × "55.84 g Fe"/(1 color(red)(cancel(color(black)("mol Fe")))) = "14.0 g Fe}$

Step 5. Calculate the percent of $\text{Fe}$ reacted

"% Fe reacted" = "mass of Fe reacted"/"original mass of Fe" × 100 % = (14.0 color(red)(cancel(color(black)("g"))))/(20 color(red)(cancel(color(black)("g")))) × 100 % = 70 %