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    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
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    A balanced equation showing the formation of iron from iron(III) oxide should be illuminating.
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    (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?
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    Fe2O6 -> 2Fe +..

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

    1:2 molar ratio
    makes sense
    sorry that was stupid haha
    thank you!
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    I much prefer your iron(III) oxide formula.
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    (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.
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    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.
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    (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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