If 4.50 l of water vapor at 50.2 °c and 0.121 atm reacts with excess iron, how many grams of iron(iii) oxide will be produced?

When the balanced equation for this reaction is:
2Fe + 3H2O → Fe2O3  +  3H2

and according to the vapour pressure formula:
PV= nRT
when we have P is the vapor pressure of H2O= 0.121 atm
and V is the volume of H2O = 4.5 L
and T in Kelvin = 52.5 +273 = 325.5 K
R= 0.08205 atm-L/g mol-K
So we can get n H2O
So, by substitution:
n H2O = PV/RT
            = (0.121*4.5)/(0.08205 * 325.5) = 0.02038 gmol
n Fe2O3 = 0.02038 * (1Fe2O3/ 3H2O) = 0.00679 gmol
Note: we get (1FeO3/3H2O) ratio from the balanced equation.
we can get the Mass of Fe2O3 from this formula:
Mass = number of moles * molecular weight       
  when we have a molecular weight of Fe2O3 = 159.7
          =  0.00679 * 159.7 = 1.084 g
∴ 1.084 gm of Fe2O3 will produced