Question

If 56 grams of `Fe_2O_3(s)` is produced, what volume in mL of `O_2` reacted at 26 deg Cand 0.95 atm?

Answer 1

Approx. `14*L`

Explanation:

`2Fe(s) + 3/2O_2(g) rarr Fe_2O_3(s)`

`"Moles of Fe"_2"O"_3=(56*g)/(159.69*g*mol^-1)=0.351*mol`.

Given the stoichiometry, `0.351*molxx3/2` `"dioxygen"` reacts, i.e. `0.526*mol`.

We assume ideality, and thus, `V=(nRT)/P`

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