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Concentration Terms - An Introduction (w/w etc)

1 mol of iron...

Question

$1 m o l$ of iron $(F e)$ reacts competely with $0.65 m o l O_{2}$ to give a mixture of only $F e O$ and $F e_{2} O_{3}$ . Mole ratio ferrous oxide to ferric oxide to ferric oxide is:

3 : 2

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4 : 3

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20 : 13

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n o n e o f t h e s e

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Solution

The correct option is B $4 : 3$
$4 : 3$

We have,

$F e + \frac{1}{2} O_{2} \to F e O$

$\Rightarrow a + 2 b = 1$ .......(1)

$2 F e + \frac{3}{2} O_{2} \to F e_{2} O_{3}$

$\Rightarrow a + 3 b = 1.3$ ......(2)

solving (1) and (2), we have

$a = 0.4, b = 0.3$

Therefore the ratio is, $4 : 3$

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Similar questions

1 mol of iron( Fe) reacts completely with 0.65 mol of O₂ to give a mixture of only FeO and Fe₂O₃ . mole ratio of ferrous oxide to ferric oxide is

Give the solution using some short cut method

1.0

0.65

O_{2}

F e O

F e_{2} O_{3} .

O_{2}

F e O

F e_{2} O_{3}

F e

O_{2}

F e O

F e_{2} O_{3}

F e

O_{2}

F e O

F e_{2} O_{3}

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