In the preparation of iron from haematite $(F e_{2} O_{3})$ by the reaction with carbon $F e_{2} O_{3} + C ⟶ F e + C O_{2}$
How much 80% pure iron could be produced from 120 kg of 90% pure $F e_{2} O_{3} ?$

94.5 kg

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60.48 kg

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116.66 kg

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120 kg

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Solution

The correct option is A 94.5 kg
Balanced reaction is
$2 F_{2} O_{3} + 3 C ⟶ 4 F e + 3 C O_{2}$
No. of moles of $F e_{2} O_{3} = (\frac{120 \times 1000}{2 \times 56 + 48}) \times \frac{90}{100}$
Mass of 80% pure iron produced
$= \frac{120 \times 1000 \times 0.9}{2 \times 56 + 48} \times \frac{2 \times 56}{0.8}$
= 94500 gram or 94.5 kg

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