When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of and . in a certain experiment, 20.00 g iron metal was reacted with 11.53 g oxygen gas. after the experiment, the iron was totally consumed, and 3.78 g oxygen gas remained. calculate the amounts of and formed in this experiment?

Given:

initial mass of Fe = 20.0 g 
initial mass of oxygen gas = 11.53 g 
mass of oxygen remained = 3.78 g 

Solution:
The chemical reaction is expressed as:

4Fe + 3O2 --> 2Fe2O3 

We first calculate the oxygen that is used in the reaction,
Mass oxygen used = 11.53 g - 3.78 g = 7.75 g O2
Moles O2 = 7.75 g O2 ( 1 mol / 32 g ) = 0.2422 moles O2
Moles Fe2O3 produced = 0.2422 moles O2 ( 2 mol Fe2O3 / 3 moles O2 ) = 0.1615 mol Fe2O3
Mass Fe2O3 produced = 0.1615 mol Fe2O3 ( 159.7 g / 1 mol ) = 25.79 g Fe2O3

Therefore, the mass of ferrous oxide produced would be 45.79 g Fe2O3.