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Percentage Composition

1.5 gm mixtur...

Question

$1.5 g m$ mixture of ${S i O}_{2}$ and ${F e}_{2} O_{3}$ on very strong heating leave a residue weighting $1.46 g m$ . The reaction responsible for loss of weight is
${F e}_{2} O_{3} (s) ⟶ {F e}_{3} O_{4} (s) + O_{2} (g)$
What is the percentage by mass of ${F e}_{2} O_{3}$ in original sample?

80 %

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20 %

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40 %

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60 %

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Solution

The correct option is B $80 %$
$3 F e_{2} O_{3} \to 2 F e_{3} O_{4} + \frac{1}{2} O_{2} (g)$
Molecular mass of $F e_{2} O_{3} = 160$
Molecular mass of $F e_{3} O_{4} = 232$
Loss in mass per $480 g m F e_{2} O_{3} = 16 g m$
In this ques, loss of mass = $0.04 g m$
Mass of $F e_{2} O_{3} = \frac{0.04}{16} \times 480$
$= 1.20 g m$
$∴ %$ mass of $F e_{2} O_{3} = \frac{1.20}{1.50} \times 100 = 80 %$

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△ H

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