MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 3, Problem 3.109P

(a)

Interpretation Introduction

Interpretation:

A balanced chemical equation for a reaction between hydrogen sulphide and oxygen to form sulphur dioxide and water vapour is to be written.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element/elements such that the number of atoms of that element(s) is the same on both sides.

Step 3: Balance the remaining atoms by placing the stoichiometric coefficients before the element/elements such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it.

Step 4: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjust the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation.

(a)

Expert Solution
Check Mark

Answer to Problem 3.109P

A balanced chemical equation is as follows:

  2H2S(g)+3O2(g)Δ2SO2(g)+2H2O(g)

Explanation of Solution

The chemical formulas for hydrogen sulphide and oxygen are H2S and O2 respectively and chemical formulas for sulphur dioxide and water vapour are SO2 and H2O respectively.

The chemical equation corresponding to the given reaction between hydrogen sulphide and oxygen to form sulphur dioxide and water vapour is :

  H2S(g)+O2(g)ΔSO2(g)+H2O(g)

Observe there are 2 O atoms in O2 on the left side of the equation and 3 O atoms in SO2 and H2O on the right side; so write 2 as the coefficient of H2O on the right and also a coefficient of 2 in front of O2 on the left side of the equation. Now this equation has a total of 4 oxygen atoms on each side and the equation becomes:

  H2S(g)+2O2(g)ΔSO2(g)+2H2O(g)

There are 4H atoms in 2H2O on the right side so put 2 as a coefficient in front of H2S on the left and the equation would become:

  2H2S(g)+2O2(g)ΔSO2(g)+2H2O(g)

There are 2S atoms as in 2H2S so a coefficient of 2 is again needed in front of SO2 on the right side of the equation and the equation becomes:

  2H2S(g)+2O2(g)Δ2SO2(g)+2H2O(g)

Now O atoms on both sides of the equation are no longer equal. There is a total of 6 O atoms on the right ( 4 in 2SO2 and 2 in 2H2O) and since there are only 4 O2 on the left side so change the coefficient from 2 to 3 in front of O2.

  2H2S(g)+3O2(g)Δ2SO2(g)+2H2O(g)

Check whether the equation is balanced or not as follows:

  Reactants (4 H, 2S, 6O) Products (4 H, 2S, 6O)

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

(b)

Interpretation Introduction

Interpretation:

A balanced chemical equation for a reaction when potassium chlorate is heated forming potassium chloride and potassium perchlorate is to be written.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element/elements such that the number of atoms of that element(s) is the same on both sides.

Step 3: Balance the remaining atoms by placing the stoichiometric coefficients before the element/elements such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it.

Step 4: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjust the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation.

(b)

Expert Solution
Check Mark

Answer to Problem 3.109P

The balanced chemical equation is as follows:

  4KClO3(s)ΔKCl(s)+3KClO4(s)

Explanation of Solution

The chemical formulas for potassium chlorate, potassium chloride, and potassium perchlorate are KClO3 , KCl and  KClO4 respectively.

The chemical equation corresponding to the given reaction when potassium chlorate  is heated forming potassium chloride and potassium perchlorate is:

  KClO3(s)ΔKCl(s)+KClO4(s)

Observe that there are 3 O atoms in KClO3 on the left side of the equation and 4 O atoms in KClO3 on the right side of it, so place 4 as a coefficient in front of KClO3 and 3 as a coefficient in front of KClO4 and the equation becomes:

  4KClO3(s)ΔKCl(s)+3KClO4(s)

Now observe that there are 12 O atoms on both sides of the equation now. Also, there are 4 Cl and 4 K atoms on both sides of the equation hence it is balanced now.

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

(c)

Interpretation Introduction

Interpretation:

A balanced chemical equation for a reaction between hydrogen gas and iron(III)oxide to form iron metal and water vapor is to be written.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element/elements such that the number of atoms of that element(s) is the same on both sides.

Step 3: Balance the remaining atoms by placing the stoichiometric coefficients before the element/elements such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it.

Step 4: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjust the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation.

(c)

Expert Solution
Check Mark

Answer to Problem 3.109P

The balanced chemical equation is as follows:

  3H2(g)+Fe2O3(s)2Fe(s)+3H2O(g)

Explanation of Solution

The chemical formulas for hydrogen gas, iron(III)oxide, iron metal and water vapor are H2, Fe2O3, Fe, H2O respectively.

The chemical equation corresponding to the given reaction is:

  H2(g)+Fe2O3(s)2Fe(s)+H2O(g)

Put a coefficient of 2 in front of iron metal on the left side of the equation as there are 2 Fe atoms in Fe2O3 and the equation would become:

  H2(g)+Fe2O3(s)2Fe(s)+H2O(g)

Next, observe that there are 3 oxygen atoms on the left so place a coefficient of 3 in front of H2O on the right side and the equation would become:

  3H2(g)+Fe2O3(s)2Fe(s)+3H2O(g)

Now there are 6 hydrogen atoms on the right so place a coefficient of 3 in front of H2 on the left side in order to make equal hydrogen atoms on both sides and the equation would appear as:

  3H2(g)+Fe2O3(s)2Fe(s)+3H2O(g)

Now check whether there is an equal number of each of the three atoms on both sides:

  Reactants (6H, 2Fe, 3O) Products (6H, 2Fe, 3O)

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

(d)

Interpretation Introduction

Interpretation:

The chemical equation for the combustion reaction of gaseous ethane to form carbon dioxide and water vapor is to be balanced.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total

mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element/elements such that the number of atoms of that element(s) is the same on both sides.

Step 3: Balance the remaining atoms by placing the stoichiometric coefficients before the element/elements such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it.

Step 4: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjust the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation.

(d)

Expert Solution
Check Mark

Answer to Problem 3.109P

The balanced chemical equation is as follows:

  2C2H6(g)+7O2(g)Δ4CO2(g)+6H2O(g)

Explanation of Solution

The chemical formulas for ethane, oxygen, carbon dioxide, and water vapor would be C2H6, O2, CO2, H2O respectively.

First write the reaction corresponding to the given combustion reaction as:

  C2H6(g)+O2(g)ΔCO2(g)+H2O(g)

There are two carbon atoms on the left side so place 2 as the coefficient for CO2 in the left side of this equation:

  C2H6(g)+O2(g)Δ2CO2(g)+H2O(g)

Next, observe that there are 6 hydrogen atoms so place 3 as a coefficient in front of H2O on the left side and equation becomes:

  C2H6(g)+O2(g)Δ2CO2(g)+3H2O(g)

Now in the right side, there are a total of 7 oxygen atoms (4 in 2CO2 and 3 in 3H2O) so since there are already 2 oxygen atoms on left side so left side requires a coefficient of 7/2 and the equation will then have seven oxygen atoms on each side as:

  C2H6(g)+72O2(g)Δ4CO2(g)+6H2O(g)        (1)

Finally in order to have only whole numbers as coefficients multiply both sides of equation (1) by 2 and equation would be:

  2C2H6(g)+7O2(g)Δ4CO2(g)+6H2O(g)

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

(e)

Interpretation Introduction

Interpretation:

The chemical equation for the reaction when iron(II)chloride is treated with chlorine trifluoride gas to form iron(III)fluoride is to be balanced.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total

mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element/elements such that the number of atoms of that element(s) is the same on both sides.

Step 3: Balance the remaining atoms by placing the stoichiometric coefficients before the element/elements such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it.

Step 4: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjust the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation.

(e)

Expert Solution
Check Mark

Answer to Problem 3.109P

The balanced chemical equation is as follows:

  2FeCl2(s)+2ClF3(g)2FeF3(s)+3Cl2(g)

Explanation of Solution

The chemical formulas for iron(II)chloride, chlorine trifluoride iron(III)fluoride and chlorine gas are FeCl2, ClF3, FeF3, Cl2 respectively.

First, write the reaction corresponding to the given reaction when iron(II)chloride is treated with chlorine trifluoride gas to form iron(III)fluoride as follows:

  FeCl2(s)+ClF3(g)FeF3(s)+Cl2(g)

Observe that there are 3 Cl atoms on the left side and 2 Cl atoms on the right side in Cl2 molecule thus a coefficient of 2 is required before Cl2 and also the coefficient of 2 before ClF3 on the left side so that the equation becomes:

  FeCl2(s)+2ClF3(g)FeF3(s)+2Cl2(g)

Next, observe that there are a total of 6 fluorine atoms (1 in FeCl2 and 3 in ClF3) on the right side of the equation so place a coefficient of 2 in front of FeF3 on the right so that the equation now has 6 Fluorine atoms on the right side too.

  FeCl2(s)+2ClF3(g)2FeF3(s)+2Cl2(g)

Now on the left side, 2 Fe atoms are present and just one in the right side so place 2 as the coefficient of FeCl2 on the left side and the equation would now become:

  2FeCl2(s)+2ClF3(g)2FeF3(s)+2Cl2(g)

Now in order to have 6 chlorine atoms on the right side change the coefficient of Cl2 on the right to 3 so that there are a total of 6 chlorine atoms on each side of the equation.

  2FeCl2(s)+2ClF3(g)2FeF3(s)+3Cl2(g)

Finally check that there are 2Fe, 6 Cl and 6 F atoms on both sides of the above equation so now this equation has been balanced.

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

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Chapter 3 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 3.1 - Prob. 3.6AFPCh. 3.1 - For many years, compounds known as...Ch. 3.1 - Use the information in Follow-up Problem 3.6A to...Ch. 3.1 - Prob. 3.7BFPCh. 3.2 - Prob. 3.8AFPCh. 3.2 - A sample of an unknown compound contains 6.80 mol...Ch. 3.2 - A sample of an unknown compound is found to...Ch. 3.2 - Prob. 3.9BFPCh. 3.2 - Prob. 3.10AFPCh. 3.2 - Prob. 3.10BFPCh. 3.2 - A dry-cleaning solvent (ℳ = 146.99 g/mol) that...Ch. 3.2 - Prob. 3.11BFPCh. 3.3 - Prob. 3.12AFPCh. 3.3 - Prob. 3.12BFPCh. 3.3 - Prob. 3.13AFPCh. 3.3 - Prob. 3.13BFPCh. 3.4 - Prob. 3.14AFPCh. 3.4 - The tarnish that forms on objects made of silver...Ch. 3.4 - Prob. 3.15AFPCh. 3.4 - In the reaction that removes silver tarnish (see...Ch. 3.4 - Prob. 3.16AFPCh. 3.4 - Prob. 3.16BFPCh. 3.4 - Prob. 3.17AFPCh. 3.4 - Prob. 3.17BFPCh. 3.4 - Prob. 3.18AFPCh. 3.4 - Prob. 3.18BFPCh. 3.4 - In the reaction in Follow-up Problem 3.18A, how...Ch. 3.4 - Prob. 3.19BFPCh. 3.4 - Prob. 3.20AFPCh. 3.4 - Prob. 3.20BFPCh. 3.4 - Marble (calcium carbonate) reacts with...Ch. 3.4 - Prob. 3.21BFPCh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Why might the expression “1 mol of chlorine” be...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Calculate the molar mass of each of the...Ch. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Calculate each of the following quantities: Mass...Ch. 3 - Calculate each of the following quantities: Amount...Ch. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Calculate each of the following: Mass % of H in...Ch. 3 - Calculate each of the following: Mass % of I in...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Oxygen is required for the metabolic combustion of...Ch. 3 - Cisplatin (right), or Platinol, is used in the...Ch. 3 - Allyl sulfide (below) gives garlic its...Ch. 3 - Iron reacts slowly with oxygen and water to form a...Ch. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - The mineral galena is composed of lead(II) sulfide...Ch. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - List three ways compositional data may be given in...Ch. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - What is the molecular formula of each...Ch. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Find the empirical formula of each of the...Ch. 3 - An oxide of nitrogen contains 30.45 mass % N. (a)...Ch. 3 - Prob. 3.45PCh. 3 - A sample of 0.600 mol of a metal M reacts...Ch. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Elemental phosphorus occurs as tetratomic...Ch. 3 - Prob. 3.76PCh. 3 - Solid iodine trichloride is prepared in two steps:...Ch. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - When 20.5 g of methane and 45.0 g of chlorine gas...Ch. 3 - Prob. 3.96PCh. 3 - Prob. 3.97PCh. 3 - Prob. 3.98PCh. 3 - Prob. 3.99PCh. 3 - Prob. 3.100PCh. 3 - Sodium borohydride (NaBH4) is used industrially in...Ch. 3 - Prob. 3.102PCh. 3 - The first sulfur-nitrogen compound was prepared in...Ch. 3 - Prob. 3.104PCh. 3 - Prob. 3.105PCh. 3 - Prob. 3.106PCh. 3 - Serotonin () transmits nerve impulses between...Ch. 3 - In 1961, scientists agreed that the atomic mass...Ch. 3 - Prob. 3.109PCh. 3 - Isobutylene is a hydrocarbon used in the...Ch. 3 - The multistep smelting of ferric oxide to form...Ch. 3 - Prob. 3.112PCh. 3 - Prob. 3.113PCh. 3 - Prob. 3.114PCh. 3 - Prob. 3.115PCh. 3 - Prob. 3.116PCh. 3 - Prob. 3.117PCh. 3 - Prob. 3.118PCh. 3 - Prob. 3.119PCh. 3 - Prob. 3.120PCh. 3 - For the reaction between solid tetraphosphorus...Ch. 3 - Prob. 3.122PCh. 3 - Prob. 3.123PCh. 3 - Prob. 3.124PCh. 3 - Prob. 3.125PCh. 3 - Prob. 3.126PCh. 3 - Prob. 3.127PCh. 3 - Prob. 3.128PCh. 3 - Ferrocene, synthesized in 1951, was the first...Ch. 3 - Prob. 3.130PCh. 3 - Prob. 3.131PCh. 3 - Citric acid (below) is concentrated in citrus...Ch. 3 - Prob. 3.133PCh. 3 - Nitrogen monoxide reacts with elemental oxygen to...Ch. 3 - Prob. 3.135PCh. 3 - Prob. 3.136PCh. 3 - Manganese is a key component of extremely hard...Ch. 3 - The human body excretes nitrogen in the form of...Ch. 3 - Aspirin (acetylsalicylic acid, C9H8O4) is made by...Ch. 3 - Prob. 3.140PCh. 3 - Prob. 3.141PCh. 3 - Prob. 3.142PCh. 3 - When powdered zinc is heated with sulfur, a...Ch. 3 - Cocaine (C17H21O4N) is a natural substance found...Ch. 3 - Prob. 3.145P
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