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"#
