Production of iron in blast furnace follows the following equation Fe _ 3 O _ 4 (s) + 4CO(g) → 3Fe(l) + 4 CO _ 2 (g) when 4 . 640kg of Fe _ 3 O _ 4 and 2 . 520kg of CO are allowed to react then the amount of iror (in g

Correct answer: C. 3360 Moles of Fe _ 3 O _ 4 = 4 . 640 × 10 ^ 3 232 = 20 Moles of CO = 2 . 52 × 10 ^ 3 28 = 90 So limiting Reagent = Fe _ 3 O _ 4 So moles of Fe formed = 60 Weight of Fe = 60 × 56 = 3360gms

JEE Main 202229 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Morning Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2022 · 29 June · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

Production of iron in blast furnace follows the following equation...

Production of iron in blast furnace follows the following equation

${Fe}_{3}{O}_{4}\left(s\right) + 4CO\left(g\right) \to 3Fe\left(l\right) + 4{CO}_{2}\left(g\right)</p><p>when$ 4 . 640kg of {Fe}_{3}{O}_{4} and 2 . 520kg of CO $are allowed to react then the amount of iror (in$ g $) produced is :$

[Given: Molar Atomic mass \left({gmol}^{- 1}\right): Fe = 56 $</p><p>Molar Atomic mass \left({gmolm}^{- 1}\right): O = 16$

Molar Atomic mass \left({gmolm}^{- 1}\right): C = 12]

Options

A $1400$
B $2200$
C $3360$
D $4200$

Correct answer

C. $3360$

Step-by-step solution

Moles of ${Fe}_{3}{O}_{4} = \dfrac{4 . 640 \times {10}^{3}}{232} = 20$

Moles of $CO = \dfrac{2 . 52 \times {10}^{3}}{28} = 90$

So limiting Reagent $= {Fe}_{3}{O}_{4}$

So moles of Fe formed $= 60$

Weight of $Fe = 60 \times 56 = 3360gms$

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