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

Question

Avery Adams · Accepted Answer

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

#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#.

Avery Adams · Answer

undefined The rust produced by oxygen gas depends on its reaction with iron. The reaction of oxygen gas with iron to form rust follows a stoichiometric ratio of 4 moles of iron for every 3 moles of oxygen gas. At room temperature and pressure, 1 mole of any ideal gas occupies approximately 22.4 liters. Therefore, 3.37 liters of oxygen gas would correspond to approximately $ \frac{3.37}{22.4} $ moles. To calculate the amount of rust produced, we need to use the stoichiometric ratio. Since the ratio is 3 moles of oxygen gas to 4 moles of iron, the amount of rust produced would be $ \frac{4}{3} \times \frac{3.37}{22.4} $ moles. You can then convert this amount into grams using the molar mass of iron oxide, which is approximately 159.69 g/mol.