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

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 $F e_{2} O_{3}$ ?

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

120 kg

60.48 kg

116.66 kg

94.5 kg

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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