In the preparation of iron from haematite (Fe2O3) by the reaction with carbon Fe2O3+C→Fe+CO2 ,how much of 80% pure iron could be produced from 120 kg of 90% pure Fe2O3 ?

2Fe2O3+3C→4Fe+3CO2 Number of moles of Fe2O3=120×1000160×90100 Mass of 80% pure iron produced =(120×1000160×90100)(2×560.8) = 94500 gm = 94.5 kg

In the preparation of iron from haematite (Fe2O3) by the reaction with carbon Fe2O3+C→Fe+CO2 ,how much of 80% pure iron could be produced from 120 kg of 90% pure Fe2O3 ?

2Fe2O3+3C→4Fe+3CO2 Number of moles of Fe2O3=120×1000160×90100 Mass of 80% pure iron produced =(120×1000160×90100)(2×560.8) = 94500 gm = 94.5 kg

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

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120 kg

60.48 kg

116.66 kg

94.5 kg

answer is A.

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Detailed Solution

$2 F e_{2} O_{3} + 3 C \to 4 F e + 3 C O_{2}$
Number of moles of $F e_{2} O_{3} = \frac{120 \times 1000}{160} \times \frac{90}{100}$
Mass of 80% pure iron produced $= (\frac{120 \times 1000}{160} \times \frac{90}{100}) (\frac{2 \times 56}{0.8})$
= 94500 gm
= 94.5 kg