A sample of 80 g of ferric oxide on reduction with coke gives 42 g iron. Percentage purity of the sample is:

100

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75

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60

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56

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Solution

The correct option is B $75$
The balanced equation for the reaction is:
$F e_{2} O_{3} + 3 C \to 2 F e + 3 C O$

160 g of $F e_{2} O_{3}$ gives $= 112 g$ of $F e$

80g of $F e_{2} O_{3}$ gives $=$ $\frac{112}{160} \times 80$ $=$ 56g of $F e$

56g of $F e$ formed from $=$ 80g of $F e_{2} O_{3}$

42g of Fe formed from $=$ $\frac{80}{56} \times 42$ $=$ 60g of $F e_{2} O_{3}$

%age purity of $F e_{2} O_{3}$ $=$ $\frac{60}{80} \times 100$ $=$ 75%

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