# Question ea867

Dec 31, 2015

$\text{m"("Al"_2"O"_3) = 55.2"g}$ $\text{to 3 significant figures}$

#### Explanation:

For the purposes of answering the question, I will assume the iron oxide in question is iron (III) oxide, ${\text{Fe"_2"O}}_{3}$, and also that this is a single displacement reaction between iron (III) oxide and aluminium metal. The equation for this reaction is as follows:

$\text{Fe"_2"O"_3 + 2"Al" -> "Al"_2"O"_3 + 2"Fe}$

Now to proceed with calculation. We must first calculate the number of moles of in the $86.4 \text{g}$ of iron (III) oxide that reacts:

"mol"("Fe"_2"O"_3) = ("m"("Fe"_2"O"_3))/("M"_r("Fe"_2"O"_3))

$\implies \frac{86.4 \text{g}}{159.6} = 0.54135$ $\text{moles to 5 significant figures}$

The mole ratio of ${\text{Fe"_2"O}}_{3}$ to ${\text{Al"_2"O}}_{3}$ is $1 : 1$, so:

$m o l \left({\text{Al"_2"O}}_{3}\right) = 0.54135$ $\text{moles to 5 significant figures}$

We can now rearrange our equation for the relationship between moles, mass (m), and relative molecular mass (${\text{M}}_{r}$) to solve for the mass of aluminium oxide:

"mol"("Al"_2"O"_3) = ("m"("Al"_2"O"_3))/("M"_r("Al"_2"O"_3)) => "m"("Al"_2"O"_3) = "mol"("Al"_2"O"_3) xx "M"_r("Al"_2"O"_3)#

$\text{m"("Al"_2"O"_3) = 0.54135 xx 102 = 55.2"g}$ $\text{to 3 significant figures}$