Question

If a piece of iron gains 10% of its weight due to partial rusting into Fe2O3? then the percentage of total iron that has rusted is?

Answer 1

The equation representing rusting of iron:

$4 Fe(s) + 3O_{2}(g) --->2Fe_{2}O_{3} (s)$

As per the stoichiometric equation,

$4 * 56 g Fe$ would react with $3 * 32 g O_{2}$

Iron gains 10 % of its mass and this excess gain is from the oxygen added.

So if we take 10 g of $O_{2}$ reacting, mass of Fe iron rusted

= $10 g O_{2} * \frac{4 * 56 g Fe}{3* 32 g O_{2}} = 23.33 g Fe$

Therefore, the percentage of total iron that has rusted = 23.33 %

Answer 2

23.33 %....................................