When 600g of iron was placed into steam, the weight of the sample became 648g. What percent of the iron reacted with the steam ?

The reaction is: 3Fe+4H2O=>Fe3O4+4H2

Update:

The answer = 21%

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  • 2 years ago
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    Stoichiometry question....

    3Fe(s) + 4H2O(g) --> Fe3O4(s) + 4H2(g)

    600.g .......................... 648g

    600.g Fe x (1 mol Fe / 55.8g Fe) x (1 mol Fe3O4 / 3 mol Fe) x (231.4g Fe3O4 / 1 mol Fe3O4) = 829.4 Fe3O4

    The theoretical yield is 829.4g Fe3O4.

    648 g / 829.4 g x 100 = 78.1% ..... which is the percent yield.

    Therefore, 78.1% of the iron reacted to make magnetite.

    ==== Or you can do it another way =====

    3Fe(s) + 4H2O(g) --> Fe3O4(s) + 4H2(g)

    600.g .......................... 648g

    648g Fe3O4 x (1 mol Fe3O4 / 231.4g Fe3O4) x (3 mol Fe / 1 mol Fe3O4) x (55.8g Fe / 1 mol Fe) = 468.8g Fe .......... the amount of Fe which reacted

    468.8 / 600.0 x 100 = 78.1% Fe reacted

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