Production of iron in blast furnace follows the following equation $$\mathrm{Fe}_{3}\mathrm{O}_{4}(\mathrm{s})+4\mathrm{CO}(\mathrm{g}) \rightarrow 3\mathrm{Fe}(\mathrm{l})+4\mathrm{CO}_{2}(\mathrm{g})$$ when 4.640 kg of $\mathrm{Fe}_{3}\mathrm{O}_{4}$ and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is : [Given : Molar Atomic mass $(\mathrm{g\,mol}^{-1})$ : Fe = 56, O = 16, C = 12]

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Some Basic Concepts of Chemistry

Production of iron in blast furnace follows the following equation

Fe_{3} O_{4} (s) + 4 CO (g) \to 3 Fe (l) + 4 CO_{2} (g)

when 4.640 kg of

Fe_{3} O_{4}

and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is :
[Given : Molar Atomic mass

(g mol^{- 1})

: Fe = 56, O = 16, C = 12]

Answer: (C)

The moles of each reactant are

\frac{4640}{232} = 20

mol of

F e_{3} O_{4}

(molar mass

3 \times 56 + 4 \times 16 = 232

) and

\frac{2520}{28} = 90

mol of

CO

(molar mass

12 + 16 = 28

). The equation consumes

4

mol of

CO

per mole of

F e_{3} O_{4}

, so

20

mol of

F e_{3} O_{4}

needs

80

mol of

CO

; since

90

mol is available,

F e_{3} O_{4}

is the limiting reactant. Each mole of

F e_{3} O_{4}

yields

3

mol of iron, giving

60

mol, so the mass of iron produced is

60 \times 56 = 3360

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option