For the purposes of answering the question, I will assume the iron oxide in question is iron (III) oxide, "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:
"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"g" of iron (III) oxide that reacts:
"mol"("Fe"_2"O"_3) = ("m"("Fe"_2"O"_3))/("M"_r("Fe"_2"O"_3))
=> (86.4"g")/159.6 = 0.54135 "moles to 5 significant figures"
The mole ratio of "Fe"_2"O"_3 to "Al"_2"O"_3 is 1 : 1, so:
mol("Al"_2"O"_3) = 0.54135 "moles to 5 significant figures"
We can now rearrange our equation for the relationship between moles, mass (m), and relative molecular mass ("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)
"m"("Al"_2"O"_3) = 0.54135 xx 102 = 55.2"g" "to 3 significant figures"
