Write down the equation:
#"Fe"_3"O"_4+"Al"rarr"Fe"_3+"Al"_2"O"_3#
Balancing, we get:
#3"Fe"_3"O"_4+8"Al"rarr3"Fe"_3+4"Al"_2"O"_3#
We need to convert both the mass of #"Fe"_3"O"_4# and #"Al"# to moles.
#(412 "g Fe"_3"O"_4)/(231.53"g/mol")=1.779"mol Fe"_3"O"_4"#
#(199 "g Al")/(26.98"g/mol")=7.376"mol Al"#
However, the iron (II, III) oxide and aluminium are in a #3:8# ratio, according to the reaction. Here, aluminium is in excess, and iron (II, III) is a limiting reagent.
We must calculate how many moles of aluminium will react.
#(3 "mol Fe"_3"O"_4)/(8 "mol Al")=(1.779"mol Fe"_3"O"_4)/(x)#
#x=(8*1.779)/(3)#
#x=4.745"mol Al"# reacts with the iron (II, III) oxide.
Since #"Al"# and #"Al"_2"O"_3# are in the ratio #8:4#, or #2:1#, we can calculate:
#(2 "mol Al")/(1 "mol Al"_2"O"_3)=(4.745"mol Al")/x#
#x=2.373 "mol Al"_2"O"_3#
We need to now calculate how many grams of aluminium oxide there will be:
#2.373 cancel("mol Al"_2"O"_3)*101.96"g/"cancel("molAl"_2"O"_3)#
#241.91"g"# of aluminium oxide is produced in the reaction.
