Question

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

Answer 1

The percent of iron reacted is 70 %.

Explanation:

Step 1. Write the chemical equation

`M_text(r):color(white)(mm)55.84color(white)(mmmmll)159.69`
`color(white)(mmmmll)"2Fe" + "3O"_2 → "Fe"_2"O"_3`
`"Mass/g:"color(white)(m)20color(white)(mmmmmmll)20`

Step 2. Calculate the moles of `"Fe"_2"O"_3`

`"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 `"Fe"`

`"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 `"Fe"`

`"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 `"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 %`