Chapter 3, Problem IISP

In a blast furnace at high temperature, iron(III) oxide in ore reacts with carbon monoxide to produce metallic iron and carbon dioxide. The liquid iron produced is cooled and weighed. The reaction is run repeatedly with the same initial mass of iron(III) oxide, 19.0 g, but differing initial masses of carbon monoxide. The masses of iron obtained arc shown in this graph.

1. (a) Write the balanced chemical equation for this reaction.
2. (b) Calculate the mass of CO required to react completely with 19.0 g iron(III) oxide.
3. (c) Calculate the mass of carbon dioxide produced when the reaction converts 10.0 g iron(III) oxide completely to products.
4. (d) From the graph, determine which reactant is limiting when less than 10.0 g carbon monoxide is available to react with 19.0 g iron(III) oxide.
5. (e) From the graph, determine which reactant is limiting when more than 10.0 g carbon monoxide is available to react with 19.0 g iron(III) oxide.
6. (f) Calculate the percent yield if 24.0 g iron(III) oxide reacted with 20.0 g carbon monoxide to produce 15.9 g metallic iron.
7. (g) Calculate the minimum mass of additional limiting reactant required to react with all of the excess of nonlimiting reactant from part (f).

Interpretation Introduction

Interpretation:

Iron $(III)$ oxide in ore reacts with carbon monoxide to produce metallic iron and carbon dioxide.

For the given reaction, balanced equation has to be written.

Concept introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Balancing the equation:

• There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• First write the skeletal reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

Explanation of Solution

Given,

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + CO (g)}\to {\text{ Fe (s) + CO}}_{\text{2}}\text{(g)}$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. There are two iron atoms in the left side and one iron atom in the right side of the reaction. Therefore, two iron atoms are added to right side of reaction. Now, the balanced equation is given below.

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + CO (g)}\to {\text{ 2Fe (s) + CO}}_{\text{2}}\text{(g)}$

Three molecule of carbon monoxide is added to the left side of the reaction and three molecules of carbon dioxide is added to the right side of the reaction. Now, the balanced equation is given below.   

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 3CO (g)}\to {\text{ 2Fe (s) + 3CO}}_{\text{2}}\text{(g)}$

Interpretation Introduction

Interpretation:

Mass of $\text{CO}$ required to react completely with $\text{19}\text{.0 g}$ iron $(III)$ oxide has to be calculated.

Explanation of Solution

Given,

    $\text{19}\text{.0 g}$ iron $(III)$ oxide

The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\text{159}\text{.688 g/mol}$.

According to the balancing equation, one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

Therefore, mass of $\text{CO}$ is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{of}\text{CO}\text{=}\text{19}\text{.0}\text{g}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\times \frac{1molF{e}_2{O}_3}{\text{159}\text{.688 g/mol}}\times \frac{3molCO}{1molF{e}_2{O}_3}\times \frac{28.0101gCO}{1molCO}\\ \text{Mass}\text{of}\text{CO}\text{=}9.998\text{g}\end{array}$

Mass of $\text{CO}$ is $\text{10}\text{.0 g}$.

Interpretation Introduction

Interpretation:

Mass of carbon dioxide produced when the reaction converts $\text{10}\text{.0 g}$ iron $(III)$ oxide has to be calculated.

Explanation of Solution

Given,

    $\text{19}\text{.0 g}$ iron $(III)$ oxide

The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\text{159}\text{.688 g/mol}$.

According to the balancing equation, one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

Therefore, mass of ${\text{CO}}_{\text{2}}$ is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{of}{\text{CO}}_{\text{2}}\text{=}\text{10}\text{.0}\text{g}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\times \frac{1molF{e}_2{O}_3}{159.688}\times \frac{3molC{O}_2}{1molF{e}_2{O}_3}\times \frac{44.0095gC{O}_2}{1molC{O}_2}\\ \text{Mass}\text{of}{\text{CO}}_{\text{2}}\text{=}8.27\text{g}\end{array}$

Mass of ${\text{CO}}_{\text{2}}$ is $\text{10}\text{.0 g}$.

Interpretation Introduction

Interpretation:

When less than $\text{10}\text{.0 g}$ carbon monoxide is available to react with $19.0g$ iron $(III)$ oxide. For the given reaction, limiting reagent has to be identified.

Concept introduction:

A limiting reactant is a reactant that is completely converted to products. Once all the limiting reactant is converted to products there is no other reactant to react.

Explanation of Solution

Given,

Figure 1

One mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 3CO (g)}\to {\text{ 2Fe (s) + 3CO}}_{\text{2}}\text{(g)}$

The given graph between the mass of Fe and mass of carbon monoxide. The graph tells that the weight of the carbon monoxide is below $\text{10 }g$, the carbon monoxide is the limiting reagent.  Therefore, the mass of the carbon monoxide is directly proportional to the mass of the compound formed. When increasing the carbon monoxide the formation of the product is more.

The calculation, shows that $\text{10}\text{.0 g}$ $\text{CO}$ is required to completely consume $19.0g$${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$. If $19.0g$${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ are present, then $\text{CO}$ will runs out first.

Interpretation Introduction

Interpretation:

When more than $\text{10}\text{.0 g}$ carbon monoxide is available to react with 19.0 g iron $(III)$ oxide. For the given reaction, limiting reagent has to be identified.

Concept introduction:

Refer to part (d)

Explanation of Solution

Given,

Figure 1

One mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 3CO (g)}\to {\text{ 2Fe (s) + 3CO}}_{\text{2}}\text{(g)}$

The given graph between the mass of Fe and mass of carbon monoxide. The graph tells that the weight of the carbon monoxide is below $\text{10 }g$, the carbon monoxide is the limiting reagent.  Therefore, the mass of the carbon monoxide is directly proportional to the mass of the compound formed. When increasing the carbon monoxide the formation of the product is more.  But after 10.0 grams of Carbon monoxide is more, ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is limiting reagent.

The graph tells that the weight of the carbon monoxide is more than $\text{10 }g$, Fe₂O₃ is limiting reagent.

The calculation, $\text{10}\text{.0 g CO}$ will be used to completely consume $\text{19}\text{.0 g}$${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$. If more than $\text{10}\text{.0 g CO}$ are present, then ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ runs out first.

Interpretation Introduction

Interpretation:

When 24.0 g iron $(III)$ oxide reacted with 20.0 g carbon monoxide to produce 15.9 g metallic iron. The percentage yield has to be calculated.

Concept introduction:

The percent yield can be calculated by using following formula

  $Percentyield=\frac{Actualmassofproduct}{Theoreticalmassofproduct}\times 100$

Explanation of Solution

Given,

    $\text{19}\text{.0 g}$ iron $(III)$ oxide

The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $159.688\text{ g/mol}$. the molar mass of $\text{Fe}$ is $55.85\text{ g/mol}$.

According to the balancing equation, one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

Given,

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

Number of moles of iron is calculated as follows,

  $\begin{array}{l}\text{Moles}\text{of}\text{Fe}\text{=}\text{24}\text{.0}{\text{g Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{× }\frac{\text{1 mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{159}\text{.688 g}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\times \frac{\text{2 mol}\text{Fe}}{\text{1 mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\\ \text{Moles}\text{of}\text{Fe}\text{=}0.301\end{array}$

Number of moles of iron is calculated as follows,

  $\begin{array}{l}\text{Moles}\text{of}\text{Fe}\text{=}\text{20}\text{.0}\text{g CO× }\frac{\text{1 mol}C\text{O}}{\text{28}\text{.0101 g}C\text{O}}\times \frac{\text{2 mol}\text{Fe}}{\text{3 mol}C\text{O}}\\ \text{Moles}\text{of}\text{Fe}\text{=}0.476\end{array}$

According to the mole calculation, ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is the limiting reagent and carbon monoxide is excess reagent.

Amount of $\text{Fe}$ produced

$\begin{array}{l}\text{ Mass}\text{of}\text{Fe}\text{=}0.301\text{mol}\text{Fe ×}\frac{\text{55}\text{.845}}{\text{1 mol}\text{Fe}}\\ \text{ Mass}\text{of}\text{Fe}\text{=}16.8\text{ g}\end{array}$

The percent yield can be calculated by using following formula

  $Percentyield=\frac{Actualmassofproduct}{Theoreticalmassofproduct}\times 100$

  $\begin{array}{l}Percentyield=\frac{15.9}{16.8}\times 100\\ Percentyield=94.6\%\end{array}$

The percent yield is $94.6\%$.

Interpretation Introduction

Interpretation:

The minimum mass of additional limiting reactant necessary to react with all of the excess of non-limiting reactant has to be calculated.

Explanation of Solution

Given,

    $\text{19}\text{.0 g}$ iron $(III)$ oxide

The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $159.688\text{ g/mol}$. the molar mass of $\text{Fe}$ is $55.85\text{ g/mol}$.

According to the balancing equation, one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

Given,

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

Number of moles of iron is calculated as follows,

  $\begin{array}{l}\text{Moles}\text{of}\text{Fe}\text{=}\text{24}\text{.0}{\text{g Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{× }\frac{\text{1 mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{159}\text{.688 g}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\times \frac{\text{2 mol}\text{Fe}}{\text{1 mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\\ \text{Moles}\text{of}\text{Fe}\text{=}0.301\end{array}$

Number of moles of iron is calculated as follows,

  $\begin{array}{l}\text{Moles}\text{of}\text{Fe}\text{=}\text{20}\text{.0}\text{g CO× }\frac{\text{1 mol}\text{CO}}{\text{28}\text{.0101 g}\text{CO}}\text{×}\frac{\text{2 mol}\text{Fe}}{\text{3 mol}\text{CO}}\\ \text{Moles}\text{of}\text{Fe}\text{=}\text{0}\text{.476}\end{array}$

According to the mole calculation, ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is the limiting reagent and carbon monoxide is excess reagent.

  $\begin{array}{l}\text{ Mass}\text{of}\text{Fe}\text{=}\text{Amount of Fe produced from reacting all the CO }\\ \text{-}\\ {\text{amount of Fe produced from reacting all the Fe}}_{\text{2}}{\text{O}}_{\text{3}}\\ \\ \text{Mass}\text{of}\text{Fe}\text{=}\text{0}\text{.476}\text{mol}\text{Fe }\text{- }\text{0}\text{.301}\text{mol}\text{Fe}\\ \\ \text{Mass}\text{of}\text{Fe}\text{=}\text{0}\text{.175}\text{mol}\text{Fe}\end{array}$

Gram of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is calculated as follows,

  $\begin{array}{l}\text{Gram}\text{of}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{=}\text{0}\text{.175}\text{mol Fe}\text{× }\frac{\text{1 mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{2}\text{mole}\text{Fe}}\text{×}\frac{\text{159}\text{.688 g}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{1 mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\\ \text{Gram}\text{of}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{=}\text{13}\text{.97}\text{g}\end{array}$

$\text{13}\text{.97 grams}$ more ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ must be added to completely react with the excess Carbon monoxide.