[Alext180]

In a single-replacement reaction such as "Iron + aluminum oxide -> _", where you are asked to predict the products, how do you know what the charge of iron is in the products? For instance the products could be 2Al + 3FeO if Iron's charge was 2+, but the products could be 2Al + 3Fe₂O if Iron's charge is just 1+, so how do you know when the charge isn't specified with Roman numerals in the reactants?

[magician4]

(...) how do you know what the charge of iron is in the products?

1. you don't
if there are several viable options  (like with iron), you'd need more information

2. the reaction type you've discussed won't happen: iron will never be able to react with Al₂O₃
it's just the other way round: Aluminium will react with Ironoxides, preferably with Fe₂O₃
Fe₂O₃ + 2 Al  2 Fe + Al₂O₃
(see: thermite reaction )

3. though being known in principal , the oxidation state of iron (+I) is rare and (mostly) irrelevant .
Furthermore, to the best of my knowledge, Fe₂O is not a known oxide of iron, as it's most instable and will react to Fe, FeO instantly

4. asides from the element, the prominent oxidation states of iron are (+II) and (+III)


regards

Ingo

[Corribus]

2. the reaction type you've discussed won't happen: iron will never be able to react with Al₂O₃

Well, it could, I suppose, if you applied an electrical bias.

[magician4]

@ Corribus

knowing a bit about the conditions Al is being produced from Al₂O3 technically, I wonder what the setup for a respective electrolysis involving Fe possibly would look like....


regards

Ingo

[Alext180]

Thanks for the reply. I guess that was a bad example because there is no reaction, how about this:

Given "silver nitrate + nickel -> _" and asked to find the products, how do you know what the oxidation number of nickel is?
So AgNO₃ + Ni -> could be Ag + NiNO₃ or Ni(NO₃)₂ etc. so how do you know?

Thanks.

[Borek]

The best approach would be to compare standard potentials to see what the Ni can be oxidized to. Naive approach tells Ni(II) is the most stable nickel form.