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.