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?

1 Answer
Jan 8, 2017

Answer:

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#