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Standard XII

Chemistry

Corrosion

If a piece of...

Question

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

23

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13

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23.3

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90

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Solution

The correct option is D $90$
Let mass of piece of iron be $m$ gram
After resulting, iron obtains= $m \times \frac{10}{100} = 0.1 m$
Number of moles before rusting= $\frac{m}{56} g$
Number of moles obtained after rusting= $\frac{0.1 m}{56}$
So, amount of mole iron that has rusted is
$\frac{m}{56} - \frac{0.1 m}{56} = \frac{0.9 m}{56}$
% of iron that has rusted= $\frac{\frac{0.9 m}{56}}{\frac{m}{56}} \times 100 = 90$ %

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

The correct option is D 90Let mass of piece of iron be m gramAfter resulting, iron obtains= m×10100=0.1mNumber of moles before rusting= m56gNumber of moles obtained after rusting= 0.1m56So, amount of mole iron that has rusted ism56−0.1m56=0.9m56% of iron that has rusted= 0.9m56m56×100=90 %

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