What mass of iron(III) oxide is required to produce $12.8 g$ of iron ? $2 F e 203 + 3 C \to 4 F e + 3 C O 2$

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Solution

Regardless of the other reactant quantities necessary, and the final products, the conservation of mass requires that we have 12.8g of Fe in the iron oxide if we expect to get at least 12.8 g of Fe metal product.
So the question is "what mass of $F e_{2} O_{3}$ contains $12. g$ Fe ?
By molar ratio, we have 2 moles of Fe per mole of $F e_{2} O_{3}$
One mole $F e_{2} O_{3}$ is $158$ grams . The two moles of Fe in comprise 111.6g of it, so we need roughly $\frac{1}{10}$ that amount.
Exactly we need $12.8 g F e \times \frac{F e_{2} O_{2} m o l e s}{111.6 g F E} \times 158 \frac{g}{m o l e} F e_{2} O_{3}$
$= 18.12 F e_{2} O_{3}$
Check
Mass ratio of $F e : F e_{2} O_{3} = \frac{111.6}{158} = 0.706$
$0.705 \times 18.12 g F e_{2} O_{3} = 12.8 g F e$

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

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

F e_{2} O_{3} (s) + 3 C (s) \to 4 Fe (s) + 3 {CO}_{2} (s)

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{F e}_{2} O_{3} + 3 C O ⟶ 2 F e + 3 C O_{2}

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