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Stoichiometric Calculations

An iron sampl...

Question

An iron sample contains $18 %$ $F e_{3} O_{4}$ . What is the amount of $F e_{3} O_{4}$ so that it is precipitated as $F e_{2} O_{3}$ which weighs $0.40$ g?

2.15

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1.075

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4.30

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2.01

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Solution

The correct option is C $2.15$ g
$F e_{3} O_{4} ⟶ 1.5 F e_{2} O_{3}$
$1 m o l 1.5 m o l$
$232 g 240 g$
$? 0.40 g$
$F e_{3} O_{4} r e q u i r e d b y 0.40 g F e_{2} O_{3} = \frac{232}{240} \times 0.4$ $= 0.3867 g (p u r e)$ $= 0.3867 \times \frac{100}{18} g$ $= 2.15 g i m p u r e F e_{3} O_{4}$ .

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