Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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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.

Chapter 3, Problem IISP, In a blast furnace at high temperature, iron(III) oxide in ore reacts with carbon monoxide to

  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).

(a)

Expert Solution
Check Mark
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,

  Fe2O3(s) + CO (g) Fe (s) + CO2(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.

  Fe2O3(s) + CO (g) 2Fe (s) + CO2(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.   

  Fe2O3(s) + 3CO (g) 2Fe (s) + 3CO2(g)

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Mass of CO required to react completely with 19.0 g iron (III) oxide has to be calculated.

Explanation of Solution

Given,

    19.0 g iron (III) oxide

The molar mass of Fe2O3 is 159.688 g/mol.

According to the balancing equation, one mole of Fe2O3 is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

Therefore, mass of CO is calculated as follows,

  MassofCO=19.0gFe2O3×1molFe2O3159.688 g/mol×3molCO1molFe2O3×28.0101gCO1molCOMassofCO=9.998g

Mass of CO is 10.0 g.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Mass of carbon dioxide produced when the reaction converts 10.0 g iron (III) oxide has to be calculated.

Explanation of Solution

Given,

    19.0 g iron (III) oxide

The molar mass of Fe2O3 is 159.688 g/mol.

According to the balancing equation, one mole of Fe2O3 is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

Therefore, mass of CO2 is calculated as follows,

  MassofCO2=10.0gFe2O3×1molFe2O3159.688×3molCO21molFe2O3×44.0095gCO21molCO2MassofCO2=8.27g

Mass of CO2 is 10.0 g.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

When less than 10.0 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:

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,

Chemistry: The Molecular Science, Chapter 3, Problem IISP , additional homework tip  1

Figure 1

One mole of Fe2O3 is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

  Fe2O3(s) + 3CO (g) 2Fe (s) + 3CO2(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 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 10.0 g CO is required to completely consume 19.0 gFe2O3. If 19.0 gFe2O3 are present, then CO will runs out first.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

When more than 10.0 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,

Chemistry: The Molecular Science, Chapter 3, Problem IISP , additional homework tip  2

Figure 1

One mole of Fe2O3 is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

  Fe2O3(s) + 3CO (g) 2Fe (s) + 3CO2(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 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, Fe2O3 is limiting reagent.

The graph tells that the weight of the carbon monoxide is more than 10 g, Fe2O3 is limiting reagent.

The calculation, 10.0 g CO will be used to completely consume 19.0 gFe2O3. If more than 10.0 g CO are present, then Fe2O3 runs out first.

(f)

Expert Solution
Check Mark
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

  Percent yield =Actual mass of productTheoretical mass of product×100

Explanation of Solution

Given,

    19.0 g iron (III) oxide

The molar mass of Fe2O3 is 159.688 g/mol. the molar mass of Fe is 55.85 g/mol.

According to the balancing equation, one mole of Fe2O3 is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

Given,

  Fe2O3(s) + 3CO(g)2Fe(s) + 3CO2(g)

Number of moles of iron is calculated as follows,

  MolesofFe=24.0g Fe2O3× 1 molFe2O3159.688 gFe2O3×2 molFe1 molFe2O3MolesofFe=0.301

Number of moles of iron is calculated as follows,

  MolesofFe=20.0g CO× 1 molCO28.0101 gCO×2 molFe3 molCOMolesofFe=0.476

According to the mole calculation, Fe2O3 is the limiting reagent and carbon monoxide is excess reagent.

Amount of Fe produced

 MassofFe=0.301molFe ×55.8451 molFe MassofFe=16.8 g

The percent yield can be calculated by using following formula

  Percent yield =Actual mass of productTheoretical mass of product×100

  Percent yield =15.916.8×100Percent yield =94.6%

The percent yield is 94.6%.

(g)

Expert Solution
Check Mark
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,

    19.0 g iron (III) oxide

The molar mass of Fe2O3 is 159.688 g/mol. the molar mass of Fe is 55.85 g/mol.

According to the balancing equation, one mole of Fe2O3 is reaction with three moles of carbon monoxide gives two moles of iron and three moles of carbon dioxide.

Given,

  Fe2O3(s) + 3CO(g)2Fe(s) + 3CO2(g)

Number of moles of iron is calculated as follows,

  MolesofFe=24.0g Fe2O3× 1 molFe2O3159.688 gFe2O3×2 molFe1 molFe2O3MolesofFe=0.301

Number of moles of iron is calculated as follows,

  MolesofFe=20.0g CO× 1 molCO28.0101 gCO×2 molFe3 molCOMolesofFe=0.476

According to the mole calculation, Fe2O3 is the limiting reagent and carbon monoxide is excess reagent.

   MassofFe=Amount of Fe produced from reacting all the CO -amount of Fe produced from reacting all the Fe2O3MassofFe=0.476molFe 0.301molFeMassofFe=0.175molFe

Gram of Fe2O3 is calculated as follows,

  GramofFe2O3=0.175mol Fe× 1 molFe2O32moleFe×159.688 gFe2O31 molFe2O3GramofFe2O3=13.97g

13.97 grams more Fe2O3 must be added to completely react with the excess Carbon monoxide.

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Chapter 3 Solutions

Chemistry: The Molecular Science

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