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Ideal Gas Equation

The number of...

Question

The number of moles of $F e_{2} O_{3}$ formed when $5.6$ of $O_{2}$ reacts with $5.6 g m$ of $F e$ is :

0.125

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0.01

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0.05

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0.1

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Solution

The correct option is C $0.05$
Balanced Reaction is:
$4 F e + 3 O_{2} \to 2 F e_{2} O_{3}$

$5.6 g m$ $5.6 g m$

$(\frac{5.6}{56})$ $(\frac{5.6}{32})$

$0.1 m o l e 0.175 m o l e$

$4 m o l e s F e$ reacts with $3 m o l e s$ of $O_{2}$ .

$∴$ 0.1 mole $F e$ reacts with $\frac{3 \times 0.1}{4} m o l e s O_{2}$

Here, Fe is limiting reagent.

4 moles of $F e$ form 2 moles of $F e_{2} O_{3}$

Mole of $F e_{2} O_{3}$ formed by 0.1 mole $F e = \frac{2}{4} \times 0.10$
$= 0.05 m o l e$

The correct option is $^{'} C^{'}$ .

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