To produce a mass of #279*g# mass of iron metal, what quantities of aluminum, and ferric oxide are required?

1 Answer
Sep 25, 2016

Answer:

Approx. #135*g# of aluminum metal, and #400*g# of iron oxide are required.

Explanation:

We need (i) a stoichometrically balanced equation:

#Fe_2O_3 + 2Al rarr Al_2O_3 + 2Fe#

And (ii) the molar quantity of iron metal produced:

#=# #(279*g)/(55.85*g*mol^-1)# #=# #5.00*mol#.

Since a #5.00*mol# of iron metal were produced, the given stoiohiometry requires at least a #2.50*mol# quantity of #"ferric oxide"#, and a #5.00*mol# quantity of #"aluminum metal"#.

Given these molar quantities, we calculate equivalent masses of #2.50*molxx159.69*g*mol^-1# #=# #??*g# of iron oxide.

And #5.00*molxx26.98*g*mol^-1# #=# #??*g# of aluminum metal.

This would be an expensive way to produce iron.