Iron III oxide is formed according to the following balanced chemical equation- 4Feg - 3O2g- 2Fe2O3gThe reaction of 8-0 moles of solid iron with 9-0 moles of oxygen gas produces -

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Iron III oxid...

Question

Iron (III) oxide is formed according to the following balanced chemical equation:
$4 F e_{(g)} + 3 O_{2 (g)} \to 2 F e_{2} O_{3 (g)}$
The reaction of $8.0$ moles of solid iron with $9.0$ moles of oxygen gas produces :

4.0 moles of

F e_{2} O_{3}

and 3.0 moles excess

O_{2}

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4.0 moles of

F e_{2} O_{3}

and 6.0 moles excess

O_{2}

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6.0 moles of

F e_{2} O_{3}

and 2.0 moles excess

O_{2}

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6.0 moles of

F e_{2} O_{3}

and 3.0 moles excess

O_{2}

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6.0 moles of

F e_{2} O_{3}

and 4.0 moles excess

O_{2}

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Solution

The correct option is A 4.0 moles of $F e_{2} O_{3}$ and 3.0 moles excess $O_{2}$
The balanced chemical equation is
$4 F e_{(g)} + 3 O_{2 (g)} \to 2 F e_{2} O_{3 (g)}$
Thus, 4 moles of Fe reacts with 3 moles of oxygen to form 2 moles of ferric oxide.
So 8 moles of Fe will react with $3 \times \frac{8}{4} = 6$ moles of oxygen to form $2 \times \frac{8}{4} = 4$ moles of ferric oxide.
$9.0 - 6 = 3.0$ moles of oxygen will remain as excess.

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Iron (III) oxide is formed according to the following balanced chemical equation:4Fe(g)+3O2(g)→2Fe2O3(g)The reaction of 8.0 moles of solid iron with 9.0 moles of oxygen gas produces :

Iron (III) oxide is formed according to the following balanced chemical equation:
$4 F e_{(g)} + 3 O_{2 (g)} \to 2 F e_{2} O_{3 (g)}$
The reaction of $8.0$ moles of solid iron with $9.0$ moles of oxygen gas produces :