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# super quick watch

1. Calculate the mass of iron (III) oxide needed to make 1g of iron.

This is what I did:
Mass = Mr Mole
Mole = Mass/Mr
Mole = 1g/ 56 (iron) = 0.018 moles
Mass of iron (III) oxide = 160*0.018 = 2.86g

The answer is 1.43g which is half... and I don't understand why?

Thanks y'all
2. A balanced equation showing the formation of iron from iron(III) oxide should be illuminating.
3. (Original post by Pigster)
A balanced equation showing the formation of iron from iron(III) oxide should be illuminating.
Fe2O3 + 3CO --> 2Fe + 3CO2

is it because there are 2 lots of Fe formed from 1 molecule of iron (III) oxide?
4. Fe2O6 -> 2Fe +..

1:2 molar ratio
5. (Original post by The Wavefunction)
Fe2O6 -> 2Fe +..

1:2 molar ratio
makes sense
sorry that was stupid haha
thank you!
6. I much prefer your iron(III) oxide formula.
7. (Original post by Pigster)
I much prefer your iron(III) oxide formula.
My "equation" just shows the reduction of the Iron and therefore the molar ratio. I don't really care about the monoxide and dioxide, it isn't important in this calculation.

The OPs is the correct, full formula.
8. Agreed, one doesn't need the full equation, just the ratio is important, but writing the formula of Fe2O3 as Fe2O6 is a little curious.
9. (Original post by Pigster)
Agreed, one doesn't need the full equation, just the ratio is important, but writing the formula of Fe2O3 as Fe2O6 is a little curious.
Oh yes, my bad. Should have been 03. Must have got the 6 from what the charge had to equal.

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