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When iron (Fe) rusts (reacts with O2), the product is iron(III) oxide (Fe2O3).
If 25 g Fe is reacted with 25 g O2 in a sealed container, what are the masses of Fe, O2 and Fe2O3 in the container after the reaction is complete?

How is type of question calculated?

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IRA

  1. substances react by molar quantities ... so you need to convert to moles

    find the moles of Fe and O2 ... by dividing by the molar masses

    two Fe and three O react to form the product

    there will be excess O2 remaining ... all the Fe will be reacted

    find the moles of Fe2O3 formed , and the moles of O2 remaining

    convert the moles to masses ... by multiplying by the molar masses

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    R_scott

  2. First, determine the reaction equation. You probably want
    4Fe + 3O2 = 2Fe2O3
    Fe is 55.8 g/mole
    O is 16 g/mole
    so,
    25g Fe = 0.448 moles
    25g O2 = 0.781 moles
    In the reaction, Fe:O2 = 4:3
    so, 0.448 moles Fe (0.224 moles of Fe2) requires 0.336 moles of O2
    So there is not enough Fe to consume all the O2. There will be 0.445 moles of O2 left over.
    So, the masses after the reaction are
    Fe: 0
    O2: 0.445*32 = 14.25g
    Fe2O3: 0.224 * 159.6 = 35.75 g
    Note that the total mass is still 50g

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    oobleck

  4. Here can you please tell me that can i use thus equation for rusting of iron Fe2O3 +3CO ----- 2Fe +3CO2

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    Faiqa

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