If 2.112 g of #Fe_2O_3# reacts with 0.687 g of #Al#, how much pure #Fe# will be produced in the reaction #Fe_2O_3(s) + 2Al(s) -> Al_2O_3(s) + 2Fe(s)#?

1 Answer
May 8, 2016

Answer:

Approx. #1.5*g# steel will be produced.

Explanation:

#"Moles of ferric oxide"=(2.112*g)/(159.69*g*mol^-1)# #=# #0.0132*mol#

#"Moles of aluminum"=(0.687*g)/(26.98*g*mol^-1)# #=# #0.0254*mol#

Aluminum is in slight deficiency, and is thus the limiting reagent. So #0.0254*molxx55.85*g*mol^-1=1.42*g#.

I should add that there is a very practical application to this reaction (which I think is still used). When setters lay train tracks, obviously they lay steel tracks over wooden or concrete sleepers. In order to join the rails between lengths they use precisely this reaction (of course they have to heat it up to get it to go!). Such welding gives a very good seal and a smooth ride.

See here