Look at your cellulose balance above and see how I did that step by step.
2Fe + O2 ==> 2FeO
I don't understand why you have trouble with that.
4Fe + 3O2 ==> 2Fe2O3
This one can be a little trickier but here is how I do it.
Fe + O2 ==> Fe2O3
I start a little backwards because since I see 2Fe on the right I KNOW I must start with a 2 on the left so I add one.
2Fe + O2 ==> Fe2O3
Now on the right I see 3 O and on the left only 2 O. How do I make up for the uneven number. The answer is since 2 and 3 won't work (3/2 would work but we can't use fractions) I just think of 6 (2*3). Actually think of it in terms of the common denominator in math problems. So I use 2 for the right and 3 for the left as follows:
Fe + 3O2 ==> 2Fe2O3
Of course that screws up the 2 Fe I had but I can change that. I now need 4 on the left since I have 4 on the right.
4Fe + 3O2 ==> 2Fe2O3
(That isn't EXACTLY what I do. I get to this point--
2Fe + O2 ==> Fe2O3 and when I see O2 on the left and O3 I think ok, how to I make O2 so it's actually 3. The answer is to multiply by 3/2 so the equation then looks like this.
2Fe + 3/2 O2 ==> Fe2O3 and it balances as follows:
2Fe on each side
3/2 O2 = 3 O on the left and 3 O on the right.
BUT we can't use fractions so how do I get rid of the 1/2 in 3/2. The answer is to multiply EVERYTHING by 2. So the above equation becomes
4Fe + 3O2 ==> 2Fe2O3.
Voila!!
I am having a hard time figuring out how to write a balanced equation for the reaction of iron and oxygen to produce iron(II) oxide.
I also need a balanced equation for the reaction of iron and oxygen to produce iron (III) oxide.
I would appreciate any and all help.
Thanking you in advance.
1 answer
1 answer