Iron rusts according to the following reaction: #4Fe (s) + 3O_2 (g) + 4H_2O -> 2Fe_2O_3(H_2O)_2#. If 1.4 liters of oxygen were added to excess water and iron at a pressure of 0.97 atms and a temperature of 55 C, how many grams of iron would rust?

1 Answer
Jun 13, 2018

Answer:

Your equation is unbalanced....

Explanation:

Let us readdress it....iron metal is oxidized to #Fe_2O_3#...

#2Fe(s) +3H_2O(l) rarr Fe_2O_3+6H^+ +6e^(-)# #(i)#

Dioxygen is reduced to water...

#1/2O_2 + 2H^+ +2e^(-) rarr H_2O(l)# #(ii)#

And #1xx(i)+3xx(ii)# gives....

#2Fe(s) +cancel(3H_2O(l)) +3/2O_2 + cancel(6H^+) +cancel(6e^(-)) rarr Fe_2O_3+cancel(6H^+) +cancel(3H_2O(l))+cancel(6e^(-))#

…. and we cancel to give...

#2Fe(s) +3/2O_2 rarr Fe_2O_3#

And note that this equation assumes COMPLETE oxidation....iron oxide chemistry, i.e. rust chemistry is a very broad church...and many iron oxides and hydrous oxides are known. And so we calculate the moles of dioxygen gas....and plug this value back into the given equation....

#n_"dioxygen"=(PV)/(RT)=(0.97*atmxx1.4*L)/(0.0821*(L*atm)/(K*mol)*328.15*K)~=0.05*mol#....and given the equation...#4/3*"equiv"# of iron metal will oxidize...approx. #4*g#..