A balanced reaction between $F e$ and $O_{2}$ to give magnetic oxide is :

2 F e + O_{2} \to 2 F e O

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2 F e + 2 O_{2} \to {F e}_{2} O_{3}

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3 F e + 2 O_{2} \to {F e}_{3} O_{4}

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F e + 2 O_{2} \to {F e}_{3} O_{4}

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Solution

The correct option is B $3 F e + 2 O_{2} \to {F e}_{3} O_{4}$
Iron is converted to iron(II, III) oxide.

$F e_{3} O_{4} i s a m a g n e t i c o x i d e .$

$3 F e + 2 O_{2} \to {F e}_{3} O_{4}$

Hence, option $(C)$ is correct.

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Similar questions

F e O (s) + {F e}_{2} O_{3} (s) ⟶ {F e}_{3} O_{4} (s)

2 F e (s) + O_{2} (g) ⟶ 2 F e O (s); Δ H^{o} = - 544 k J

4 F e (s) + 3 O_{2} (g) ⟶ 2 {F e}_{2} O_{; 3} (s); Δ H^{o} = - 1648.4 k J

{F e}_{3} O_{4} (s) ⟶ 3 F e (s) + 2 O_{2} (g); Δ H^{o} = + 1118.4 k J

2 F e + \frac{3}{2} O_{2} \to F e_{2} O_{3}; x k J / m o l

2 F e + O_{2} \to 2 F e O; y k J / m o l

Δ H

F e_{3} O_{4}

O_{2}

F e

F e O

{F e}_{2} O_{3}

{F e}_{3} O_{4}

C u + H_{2} S O_{4} \to C u S O_{4} + {S O}_{2} ↑ + H_{2} O

F e + H_{2} O \to {F e}_{3} O_{4} + H_{2} ↑

{C H}_{4} + O_{2} \to C O_{2} + H_{2} O

F e + H_{2} O \to {F e}_{3} O_{4} + H_{2} ↑

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