For the reaction ${F e}_{2} O_{3} + 3 C O ⟶ 2 F e + 3 C O_{2}$ the volume of carbon monoxide required to reduce one mole of ferric oxide is:

22.4 {d m}^{3}

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44.8 {d m}^{3}

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67.2 {d m}^{3}

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11.2 {d m}^{3}

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Solution

The correct option is C $67.2 {d m}^{3}$
Based on the given equation it is clear that $1$ mole of $F e_{2} O_{3}$ i.e ferric oxide is reduced by $3$ carbon monoxide.

$1$ mole of any compound occupies $22.4 l$ of volume.

So, the volume of carbon monoxide required to reduce one mole of ferric oxide is $= 3 \times 224.4 d m^{3} = 67.2 d m^{3}$ .

Hence, option C is correct.

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Similar questions

F e_{2} O_{3}

F e_{2} O_{3} + 3 C O \to 2 F e + 3 C O_{2}

F e_{2} O_{3} + 3 C O \to 2 F e + 3 C O_{2}

F e_{2} O_{3} + 3 C O \to 2 F e + 3 C O_{2}

F e_{2} O_{3} + 3 C O \to 2 F e + 3 C O_{2}

{F e}_{2} O_{3} + 3 C O \to 2 F e + 3 {C O}_{2} F e O + C O \to F e + {C O}_{2} △ r_{H} = - 26 k J △ r_{H} = - 16 k J

△ H

{F e}_{2} O_{3} + C O \to 2 F e O + {C O}_{2} △ r_{H} = ?

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