Final thermochemistry question?

Question

The thermite reaction, used for welding iron, is the reaction of Fe3O4 with Al.

8Al (s) + 3Fe3O4 (s) = 4Al2O3 (s) + 9Fe (s)
H = -3350jk/mol

Because this large amount of heat cannot be rapidly dissipated to the surroundings, the reacting mass may reach temperatures near 3000C. How much heat is released by the reaction of 23g of Al with 57.6g of Fe3O4.


Im lost here. I was told i needed to find the limiting reactant? why? overall im totally lost here.

thank you.

Answer 1

Atomic weights: Al=27 Fe=56 Fe3O4=232

23gAl x 1molAl/27gAl =0.852 mole Al

57.6gFe3O4 x 1molFe3O4/232gFe3O4 = 0.248 mole Fe3O4

0.852 mole Al requires (3/8)(0.852) = 0.320 mole Fe3O4, and you don't have that much, so Fe3O4 is limiting. Work with that.

Answer 2

You need to find the limiting reagent in order to determine how much Al and Fe3O4 react, as they react in a 3 to 8 molar ratio. So, . . .

(1) convert Al and Fe3O4 to moles: 23/26.9 = 0.855 and 57.6/231.4 = 0.249
(2) divide each mole amount by the coefficient: 0.855/8 and 0.249/3
(3) The lower value from step 2 tells us which is the limiting reagent; it is the Fe3O4

Now, we know that the reaction stopped after 0.249 moles of Fe3O4 was used.

Your enthalpy value is a bit wrong. it is actually -3350 kJ for the entire reaction as written. When 3 moles of Fe3O4 is consumed, 3350 kJ is given off.

So, do this 3350 kJ is to 3 moles as x is to 0.249 moles.

This will give you the kJ released when 0.249 moles of Fe3O4 reacts. make sure to put a negative sign on it.