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The number of moles of $F e_{2} O_{3}$ formed when 56 lit of $O_{2}$ reacts with 56 gm of $F e$

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1.25

0.1

0.5

1.0

answer is C.

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Detailed Solution

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

$4 \times 56 g m F e \overset{}{\to} 3 \times 22.4 l i t O_{2}$ $F e$

L.R $= F e$

$4 \times 56 g m F e$ $\to_{}^{}$ 2 mole $F e_{2} O_{3}$

56 gm $F e$ $\to_{}^{}$ ?

$= \frac{2}{4 \times 56} \times 56 = 0.5$ mole

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