Question

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of FeO and Fe₂O₃. In a certain experiment, 20.00 g of iron metal was reacted with 11.20 g of oxygengas. After the experiment, the iron was totally consumed and 3.24 g of oxygen gas remained. Calculate the amounts of FeO and Fe₂O₃formed in this experiment.

Short answer

The amounts of FeOand Fe₂O₃ formedin this experiment is 5.75g and 22.26g.

Step-by-step solution

Definition

The molar mass of a materialis defined as the mass of the substancecontained in one moleof the substance, and it is equal to the total massof the component atoms. It may be stated mathematically as follows:
Molar mass$=\frac{Weighting}{numberofmoles}$

Determination of mass of FeO and Fe2O3

From the given, the equations are

1. $2Fe+O_2\to 2FeO$
   
2. $4Fe+3O_2\to 2F{e}_2{O}_3$
   

$\frac{20g.z}{55.845{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mole$}\right.}}\times \frac{1mole{O}_2}{2moleFe}=0.1791=zmole{O}_2$in equation (1)

$\frac{20g.(1-z)}{55.845{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mole$}\right.}}\times \frac{3mole{O}_2}{4moleFe}=0.268601-0.268601=zmole{O}_2$in equation (2)

$n{\left(O_2\right)}_{reacted}=\frac{11.20-3.42}{32\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mole$}\right.}=0.24875mole$

Therefore,z = 0.22171mole

$$\begin{gathered} \frac{20G.0.22171mole}{55.845{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mole$}\right.}}\times \frac{2moleFeO}{2moleFe}\times 71.844\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mole$}\right.FeO=5.705g \\ \frac{20G.(1-0.22171mole)}{55.845{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mole$}\right.}}\times 4\frac{2moleF{e}_2{O}_3}{2moleFe}\times 159.6882\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mole$}\right.F{e}_2{O}_3=22.26g \end{gathered}$$

Therefore, m(FeO) = 5.75g and (Fe₂O₃) = 22.26g.