Chapter 11, Problem 103A

Iron reacts with oxygen as Shown.

   $\text{4Fe}\left(\text{s}\right){\text{+3O}}_{\text{2}}\left(\text{g}\right)\to {\text{2FeO}}_{\text{3}}\left(\text{s}\right)$

Different amounts of iron were burned in a fixed amount of oxygen. For each mass of iron burned, the mass Of iron(lll) oxide formed was plotted on the graph shown Figure 11.15. Why does the graph level off after 25.0 g of iron is burned? How many moles of oxygen are present in the fixed amount?

Interpretation Introduction

Interpretation:

In the given graph, the reason for the level of the graph to be off after 25 g of iron needs to be explained. The number of moles of oxygen present in a fixed amount needs to be determined.

Concept Introduction: Iron reacts with oxygen to produce iron oxide.

Explanation of Solution

A linear increase in the graph suggests Fe₂ O₃ was produced according to the mass of iron. Up to 25 g of iron is present in fewer amounts. But, after that, oxygen which is present in a fixed amount is not enough to convert the remaining amount of iron to iron oxide. After this point, oxygen acts as a limiting reactant. This is the reason why there is an off on the graph level.

At point 25 g, iron oxygen is present in a sufficient amount. So at this point, the number of moles of oxygen can be calculated.

For that

Step 1: Write a balanced chemical equation

  $\begin{array}{l}\text{4Fe(s)}+{\text{3O}}_2\text{(g)}\to {\text{2Fe}}_2{\text{O}}_3\text{(s)}\end{array}$

  $\begin{array}{c}\text{25 g}\text{? g}\end{array}$

Step 2: Convert the mass of reactant to moles

  $\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{Fe}=\text{55}\text{.84}\text{g/mole}\end{array}$

       $\begin{array}{c}\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\end{array}$

  $\begin{array}{c}\text{Moles}\text{of}\text{Fe=}\frac{\text{25}}{\text{55}\text{.84}}\text{=0}\text{.447}\text{moles}\end{array}$

Step 3: Convert moles of iron to moles of oxygen required using mole ratio

4 mole of Fe require = 3 moles of O₂

1 mole of Fe requires $\frac{3}{4}$ moles of O₂

0.447 mole of Fe requires = $\frac{3}{4}\times 0.447$ moles of O₂ = 0.335 moles of O₂

Step 4: Convert moles of oxygen to mass

       $\begin{array}{c}\text{Molar}\text{}\text{mass}\text{of}{\text{O}}_2=\text{16g/mole}\end{array}$

             $\begin{array}{c}\text{Mass}=\text{Molar}\text{mass×Moles}\end{array}$

  $\begin{array}{c}\text{Mass}\text{of}{\text{H}}_2\text{(Theoretical}\text{yield)}=\text{16×0}\text{.335}\end{array}$

                $\begin{array}{c}=\text{5}\text{.36g}\end{array}$