Ferric oxide can be obtained by oxidation of $FeO :$
$4 FeO + O_{2} ⟶ 2 {Fe}_{2} O_{3}$ ..(i)
The $O_{2}$ gas required can be prepared by the following reaction.
$2 {SO}_{3} ⟶ 2 {SO}_{2} + O_{2} (g)$
What is the maximum amount of ${Fe}_{2} O_{3}$ that can be produced by $144 gFeO$ and $160 {gSO}_{3}$ ? [Atomic mass of $Fe = 56$ ]

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$160 g$

$320 g$

$80 g$

$120 g$

answer is D.

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Detailed Solution

$n_{O_{2}}$ from equation $(ii) = \frac{160}{80} \times \frac{1}{2} = 1$ mol

$n_{FeO} = 2 mole$

$F e O$ is $L R$ in equation (i)

$W_{{Fe}_{2} O_{3}} = 1 \times 160 = 160 g$

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