If a piece of iron gains 10% of its weight due to partial rusting into $F e_{2} O_{3}$ , the percentage of total iron that has rusted is:

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23.3

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Solution

The correct option is C 23.3
Let the total mass of $F e = 100 g$
So, the mass of $O_{2} = 10 g$
$2 F e + \frac{3}{2} O_{2} \to F e_{2} O_{3}$
From the stoichiometry of the reaction,
$\frac{10}{32}$ mol of $O_{2}$ will combine with $\frac{10}{24}$ mol of $F e$
Mass of $F e = \frac{10}{24} \times 56 = 23.3 g$ or 23.3%

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

The correct option is C 23.3Let the total mass of Fe=100 g So, the mass of O2=10 g 2Fe+32O2→Fe2O3 From the stoichiometry of the reaction, 1032 mol of O2 will combine with 1024 mol of Fe Mass of Fe=1024×56=23.3 g or 23.3%

If a piece of iron gains 10% of its weight due to partial rusting into $F e_{2} O_{3}$ , the percentage of total iron that has rusted is: