How much rust can be produced by a #3.37*L# volume of oxygen gas at room temperature and pressure?

1 Answer
anor277
Jan 9, 2018

Well, we can give one scenario.....

Explanation:

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

....and here I have proposed ferric oxide as the product....there are other iron oxides..and clearly we will have to solve the other options, #FeO#, #Fe_3O_4# etc. individually.

We assess the molar quantity of dioxygen gas....

#n=(PV)/(RT)=(1*atmxx3.37*L)/(0.0821*(L*atm)/(mol*K)*298*K)=0.138*mol#...and we require #4/3# equiv with respect to iron....i.e. #0.184*molxx55.8*g*mol^-1=10.25*g#.

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