Calculate the percentage composition of iron in ferric oxide, $F e_{2} O_{3}$ .
[Atomic mass of $F e = 56$ ].

70%

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60%

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50%

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40%

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Solution

The correct option is A 70%
$F e_{2} O_{3}$

$1$ Fe atomic mass = $56$

$1$ oxygen atomic mass= $16$

total = $56 \times 2 + 16 \times 3$

$= 160$

$% F e$ = $\frac{56 \times 2}{160} \times 100$

$= 70 %$

Hence, the correct option is $A$

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