On rusting of iron it completely converts in $F e_{2} O_{3}$ . If a sample of pure gains $10 %$ in weight due to partial rusting. Calculate the fraction of iron rusted.

5 / 30

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8 / 30

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7 / 30

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9 / 30

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Solution

The correct option is A $7 / 30$
Let, initially sample has $x$ gm weight in which $y$ gm is $F e$ .
For every $F e$ that changes to $F e_{2} O_{3}$ , $(3 / 2) \times 16$ gm weight is gained.
No. of $F e$ in $x$ gm sample $= (y / 56)$
therefore, increase in weight for $y / 56$ $F e = (3 / 2) \times 16 \times (y / 56)$ .
% of $F e$ in sample $= \frac{7}{30}$

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If rusting of iron is a chemical change than how it converts back to its original form (iron). As I have seen that if we pore cold drinks on rusting iron and keep it for sometimes. And then if we wash the iron with water than it will convert back to its original form. I have seen with my own eyes......

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On rusting of iron it completely converts in Fe2O3. If a sample of pure gains 10% in weight due to partial rusting. Calculate the fraction of iron rusted.

The correct option is A 7/30Let, initially sample has x gm weight in which y gm is Fe.For every Fe that changes to Fe2O3 , (3/2)×16 gm weight is gained.No. of Fe in x gm sample =(y/56)therefore, increase in weight for y/56 Fe=(3/2)×16×(y/56).% of Fe in sample =730

On rusting of iron it completely converts in $F e_{2} O_{3}$ . If a sample of pure gains $10 %$ in weight due to partial rusting. Calculate the fraction of iron rusted.