Question

An ore contains Fe3O4 and no other iron. The iron in a 56.7-gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 26.7 g. What was the mass of Fe3O4 in the sample of ore?

Answer 1

$Fe_3O_4$ is converted to $Fe_2O_3$ by the following reaction:

$4Fe_3O_4+O_2 \to 6 Fe_2O_3$

As per the stoichiometry of the reaction,

6 moles of $Fe_2O_3$ is equivalent to 4 moles of $Fe_3O_4$

or, $6 \times (56 \times 2 +16 \times 3) \space gm \space Fe_2O_3 \equiv$

$4 \times(56 \times 3 +16 \times 4)\space gm \space Fe_3O_4$

or,$6 \times 160 \space gm Fe_2O_3 \equiv 4 \times 232 \space gm \space Fe_3O_4$

or,$26.7 \space gm \space Fe_2O_3\equiv\frac{4 \times 232}{6 \times 160}\times26.7 \space gm \space Fe_3O_4$

or, 26.7 grams of $Fe_2O_3$ will have 25.81 grams of $Fe_3O_4$ in the ore sample.