If 56 grams of $F e_{2} O_{3} (s)$ $Fe_2O_3(s)$ is produced, what volume in mL of $O_{2}$ $O_2$ reacted at 26 deg Cand 0.95 atm?

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If 56 grams of $F e_{2} O_{3} (s)$ $Fe_2O_3(s)$ is produced, what volume in mL of $O_{2}$ $O_2$ reacted at 26 deg Cand 0.95 atm?

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Answer 1 · 2017-01-08T15:38:39

Approx. $14 \cdot L$ $14*L$

Explanation:

$2 F e (s) + \frac{3}{2} O_{2} (g) \to F e_{2} O_{3} (s)$ $2Fe(s) + 3/2O_2(g) rarr Fe_2O_3(s)$

${Moles of Fe}_{2} O_{3} = \frac{56 \cdot g}{159.69 \cdot g \cdot m o l^{- 1}} = 0.351 \cdot m o l$ $"Moles of Fe"_2"O"_3=(56*g)/(159.69*g*mol^-1)=0.351*mol$ .

Given the stoichiometry, $0.351 \cdot m o l \times \frac{3}{2}$ $0.351*molxx3/2$ $dioxygen$ $"dioxygen"$ reacts, i.e. $0.526 \cdot m o l$ $0.526*mol$ .

We assume ideality, and thus, $V = \frac{n R T}{P}$ $V=(nRT)/P$

$=(0.526*cancel(mol)xx0.0821*(L*cancel(atm))/(cancel(K)*cancel(mol))xx299*cancel(K))/(0.95*cancel(atm))=??L$

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If 56 grams of $F e_{2} O_{3} (s)$ $Fe_2O_3(s)$ is produced, what volume in mL of $O_{2}$ $O_2$ reacted at 26 deg Cand 0.95 atm?

1 Answer

Explanation:

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If 56 grams of $F e_{2} O_{3} (s)$ $Fe_2O_3(s)$ is produced, what volume in mL of $O_{2}$ $O_2$ reacted at 26 deg Cand 0.95 atm?