Fe+S8 =>?? Predicting products
what charge is the iron present (II or III)?
Assuming iron's charge is (II), and that you mean 8S for an easier time balancing with S. Then the reaction of heating up iron + sulfur would be: 8Fe + 8S --> 8FeS but simply coefficients would just be: Fe + S --> FeS
@Sheng Thanks! I have another question. Do you mind answering?
go for it
In the equation, Na+N2, it predicts that Na3N is its unbalanced product. Why is there only one N in the product when there is 2 in the first equation? @Sheng
because it is "unbalanced"
What do you mean? @Sheng
What I mean is, when you first solve the equation, how do you come up with Na3N before balancing it?
Oh, nitrogen has an oxidation number of -3, while sodium has an oxidation number of +1 to make a neutral compound of sodium and nitrogen, you need 3 sodiums (+1x3) to balance with 1 nitrogen (-3) 3-3 = 0.
But how do you come up with 1 N in Na3N? @Sheng
because the compound doesnt need another nitrogen, it's already satisfied the octet and is a neutral compound
Why do you get rid of the 2 from the N on the original equation? What happens to the 2?
ok to make this simpler to understand lets just take your unbalanced original equation and balance it: Na + N2 --> Na3N (unbalanced) 2 nitrogens on 1 side, 1 on the other, 3sodiums on 1 side, 1 on the other 6Na + N2 --> 2Na3N here you see, 6 sodiums on both sides, 2 nitrogens on both sides
Sorry for the stupid questions. This is really hard for me
@Sheng So the original equation already makes Nitrogen diatomic?
nitrogen, oxygen, hydrogen are all diatomic elements they will always exist in a diatomic form
Okay so the original equation already takes care of making it a diatomic ?
@Sheng
yes you'll only ever see N2 in equations when talking about just nitrogen
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