when the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of FeO and a certain experiment, 20.00g of iron metal was reacted with 11.20g of oxygen gas. After the experiment the iron was totally consumed and 3.56g oxygen gas remained. Calculate the amounts of FeO and Fe 203 formed in this experiment

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  1. This is another set of simultaneously solved equations but it is more complicated.
    You must separate, somehow, the Fe used to form FeO from the Fe used to form Fe2O3 and/or separate the oxygen used for form FeO and oxygen used to form Fe2O. Then solve stoichiometry problem to find how much FeO and Fe2O3 were formed.
    2 Fe + O2 ==> 2FeO
    4Fe + 3O2 ==> 2Fe2O3.
    Fe used = 20.00 g
    O2 used = 11.20-3.56 = 7.64 g

    Let X = g Fe used to form FeO
    and Y = g Fe used to form Fe2O3 so
    eqn 1 is X + Y = 20.00

    For equation 2 I used O2 needed for the X part of the 20g Fe + O2 needed for the Y part of the 20 g Fe. That is
    (X/2*55.85) + (3Y/4*55.85) = 7.65/32

    When you find X and Y, the use regular stoichiometry to convert X to grams FeO formed and Y to grams Fe2O3 formed. Post your work if you get stuck.

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  2. The first line of the question states that a mixture of iron II oxide and ironIII oxide is formed WHEN oxygen is limited. But in this situation Oxygen is in excess (it starts with 11.20g of Oxygen and end with 3.56g Oxygen while the metal is used up). So only one form of oxide will preferentially be formed. That’s how I look at it.

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