Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 9, Problem 18CR
Interpretation Introduction

(a)

Interpretation:

The amount of product in given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

2 PbS +  3 O22PbO +  2SO2  Reactants            Products

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Expert Solution
Check Mark

Answer to Problem 18CR

There are 5.3 g SiCl4 and 0.375 g C formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

SiC(s)+2Cl2(g)SiCl4(l)+C(s)

Given:

Amount of first reactant = 12.5 g

Calculation:

Number of moles of SiC calculated as follows:

Number of moles=mass in gmolarmass=1.25 g40.11 g/mol=0.0312 moles SiC

Amount of product in gram calculated as follows:

0.0312 moles SiC×1.00 mole SiCl41.00 moles SiC×169.9 g SiCl41.00 mole SiCl4=5.3 g SiCl4

0.0312 moles SiC×1.00 mole C1.00 moles SiC×12.01 g C1.00 mole C=0.375 g C.

Interpretation Introduction

(b)

Interpretation:

The amount of product in the given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

2 PbS +  3 O22PbO +  2SO2  Reactants            Products

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Expert Solution
Check Mark

Answer to Problem 18CR

There is 2.00 g LiOH formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

Li2O(s)+H2O(l)2LiOH(aq)

Given:

Amount of first reactant = 12.5 g

Calculation:

Number of moles of Li2O calculated as follows:

Number of moles=mass in gmolarmass=1.25 g29.88 g/mol=0.0418 moles Li2O

Amount of product in gram calculated as follows:

0.0418 moles Li2O×2.00 mole LiOH1.00 moles Li2O×23.95 g LiOH1.00 mole LiOH=2.00 g LiOH.

Interpretation Introduction

(c)

Interpretation:

To calculate the amount of product in given reaction

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

2 PbS +  3 O22PbO +  2SO2  Reactants            Products

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Expert Solution
Check Mark

Answer to Problem 18CR

There are 2.56 g NaOH and 0.512 g O2 formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

2Na2O2(s)+2H2O(l)4NaOH(aq)+O2(g)

Given:

Amount of first reactant = 12.5 g

Calculation:

Number of moles of Na2O2 calculated as follows:

Number of moles=mass in gmolarmass=1.25 g77.98 g/mol=0.016 molesNa2O2

Amount of product in gram calculated as follows:

0.016 molesNa2O2×4.00 mole NaOH1.00 molesNa2O2×39.997  g NaOH1.00 mole NaOH=2.56 g NaOH0.016 molesNa2O2×1.00 mole O21.00 molesNa2O2×32.00  g O21.00 mole O2=0.512 g O2.

Interpretation Introduction

(d)

Interpretation:

The amount of product in given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

2 PbS +  3 O22PbO +  2SO2  Reactants            Products

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Expert Solution
Check Mark

Answer to Problem 18CR

There are 0.984 g Sn and 0.299 g H2O formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

SnO2(s)+2H2(g)Sn(s)+2H2O(l)

Given Information:

Amount of first reactant = 12.5 g

Calculation:

Number of moles of SnO2 calculated as follows:

Number of moles=mass in gmolarmass=1.25 g150.71 g/mol=0.00829 moles SnO2

Amount of product in gram calculated as follows:

0.00829 moles SnO2×1.00 mole Sn1.00 moles SnO2×118.71 g Sn1.00 mole Sn=0.984 g Sn0.00829 moles SnO2×2.00 mole H2O1.00 moles SnO2×18.02 g H2O1.00 mole H2O=0.299 g H2O.

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Chapter 9 Solutions

Introductory Chemistry: A Foundation

Ch. 9 - Nitrogen (N2) and hydrogen (H2)react to form...Ch. 9 - Prob. 4ALQCh. 9 - ou know that chemical A reacts with chemical B....Ch. 9 - f 10.0 g of hydrogen gas is reacted with 10.0 g of...Ch. 9 - Prob. 7ALQCh. 9 - Prob. 8ALQCh. 9 - hat happens to the weight of an iron bar when it...Ch. 9 - Prob. 10ALQCh. 9 - What is meant by the term mole ratio? Give an...Ch. 9 - Which would produce a greater number of moles of...Ch. 9 - Consider a reaction represented by the following...Ch. 9 - Prob. 14ALQCh. 9 - Consider the balanced chemical equation...Ch. 9 - Which of the following reaction mixtures would...Ch. 9 - Baking powder is a mixture of cream of tartar...Ch. 9 - You have seven closed containers each with equal...Ch. 9 - Prob. 19ALQCh. 9 - Prob. 20ALQCh. 9 - Consider the reaction between NO(g)and...Ch. 9 - hat do the coefficients of a balanced chemical...Ch. 9 - he vigorous reaction between aluminum and iodine...Ch. 9 - Prob. 3QAPCh. 9 - hich of the following statements is true for the...Ch. 9 - or each of the following reactions, give the...Ch. 9 - or each of the following reactions, give the...Ch. 9 - Prob. 7QAPCh. 9 - Prob. 8QAPCh. 9 - onsider the balanced chemical equation...Ch. 9 - Write the balanced chemical equation for the...Ch. 9 - For each of the following balanced chemical...Ch. 9 - Prob. 12QAPCh. 9 - For each of the following balanced chemical...Ch. 9 - For each of the following balanced chemical...Ch. 9 - For each of the following unbalanced equations,...Ch. 9 - For each of the following unbalanced equations,...Ch. 9 - What quantity serves as the conversion factor...Ch. 9 - Prob. 18QAPCh. 9 - Using the average atomic masses given inside the...Ch. 9 - Using the average atomic masses given inside the...Ch. 9 - Using the average atomic masses given inside the...Ch. 9 - Using the average atomic masses given inside the...Ch. 9 - For each of the following unbalanced equations,...Ch. 9 - For each of the following unbalanced equations,...Ch. 9 - For each of the following unbalanced equations,...Ch. 9 - Boron nitride reacts with iodine monofluoride i...Ch. 9 - “Smelling salts,” which are used to revive someone...Ch. 9 - Calcium carbide, CaC2, can be produced in an...Ch. 9 - When elemental carbon is burned in the open...Ch. 9 - If baking soda (sodium hydrogen carbonate) is...Ch. 9 - Although we usually think of substances as...Ch. 9 - When yeast is added to a solution of glucose or...Ch. 9 - Sulfurous acid is unstable in aqueous solution and...Ch. 9 - Small quantities of ammonia gas can be generated...Ch. 9 - Elemental phosphorus bums in oxygen with an...Ch. 9 - Prob. 36QAPCh. 9 - Ammonium nitrate has been used as a high explosive...Ch. 9 - If common sugars arc heated too strongly, they...Ch. 9 - Thionyl chloride, SOCl2, is used as a very...Ch. 9 - Prob. 40QAPCh. 9 - Which of the following statements is(are) true? l...Ch. 9 - Explain how one determines which reactant in a...Ch. 9 - Consider the equation: 2A+B5C. If 10.0 g of A...Ch. 9 - Balance the following chemical equation, and then...Ch. 9 - For each of the following unbalanced reactions,...Ch. 9 - For each of the following unbalanced chemical...Ch. 9 - For each of the following unbalanced chemical...Ch. 9 - For each of the following unbalanced chemical...Ch. 9 - For each of the following unbalanced chemical...Ch. 9 - For each of the following unbalanced chemical...Ch. 9 - Lead(II) carbonate, also called “white lead,” was...Ch. 9 - Copper(II) sulfate has been used extensively as a...Ch. 9 - Lead(II) oxide from an ore can be reduced to...Ch. 9 - If steel wool (iron) is heated until it glows and...Ch. 9 - A common method for determining how much chloride...Ch. 9 - Although many sulfate salts are soluble in water,...Ch. 9 - Hydrogen peroxide is used as a cleaning agent in...Ch. 9 - Silicon carbide, SIC, is one of the hardest...Ch. 9 - Prob. 59QAPCh. 9 - The text explains that one reason why the actual...Ch. 9 - According to his prelaboratory theoretical yield...Ch. 9 - An air bag is deployed by utilizing the following...Ch. 9 - The compound sodium thiosutfate pentahydrate....Ch. 9 - Alkali metal hydroxides are sometimes used to...Ch. 9 - Although they were formerly called the inert...Ch. 9 - Solid copper can be produced by passing gaseous...Ch. 9 - Prob. 67APCh. 9 - Prob. 68APCh. 9 - Prob. 69APCh. 9 - When the sugar glucose, C6H12O6, is burned in air,...Ch. 9 - When elemental copper is strongly heated with...Ch. 9 - Barium chloride solutions are used in chemical...Ch. 9 - The traditional method of analysis for the amount...Ch. 9 - For each of the following reactions, give the...Ch. 9 - Prob. 75APCh. 9 - Consider the balanced equation...Ch. 9 - For each of the following balanced reactions,...Ch. 9 - For each of the following balanced equations,...Ch. 9 - Prob. 79APCh. 9 - Using the average atomic masses given inside the...Ch. 9 - For each of the following incomplete and...Ch. 9 - Prob. 82APCh. 9 - Prob. 83APCh. 9 - It sodium peroxide is added to water, elemental...Ch. 9 - When elemental copper is placed in a solution of...Ch. 9 - When small quantities of elemental hydrogen gas...Ch. 9 - The gaseous hydrocarbon acetylene, C2H2, is used...Ch. 9 - For each of the following unbalanced chemical...Ch. 9 - For each of the following unbalanced chemical...Ch. 9 - Hydrazine N2H4, emits a large quantity of energy...Ch. 9 - Consider the following reaction:...Ch. 9 - Before going to lab, a student read in his lab...Ch. 9 - Consider the following unbalanced chemical...Ch. 9 - Prob. 94CPCh. 9 - Consider the following unbalanced chemical...Ch. 9 - Over the years, the thermite reaction has been...Ch. 9 - Consider the following unbalanced chemical...Ch. 9 - Ammonia gas reacts with sodium metal to form...Ch. 9 - Prob. 99CPCh. 9 - he production capacity for acrylonitrile (C3H3N)in...Ch. 9 - Prob. 1CRCh. 9 - erhaps the most important concept in introductory...Ch. 9 - ow do we know that 16.00 g of oxygen Contains the...Ch. 9 - Prob. 4CRCh. 9 - hat is meant by the percent composition by mass...Ch. 9 - Prob. 6CRCh. 9 - Prob. 7CRCh. 9 - Prob. 8CRCh. 9 - Prob. 9CRCh. 9 - Consider the unbalanced equation for the...Ch. 9 - Prob. 11CRCh. 9 - What is meant by a limiting reactant in a...Ch. 9 - Prob. 13CRCh. 9 - Prob. 14CRCh. 9 - Prob. 15CRCh. 9 - Prob. 16CRCh. 9 - A compound was analyzed and was found to have the...Ch. 9 - Prob. 18CRCh. 9 - Prob. 19CRCh. 9 - Solid calcium carbide (CaC2)reacts with liquid...Ch. 9 - A traditional analysis for samples containing...
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  • A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed of only N2 and O2 with a molar ratio of 3.76:1.00, and the N2 remains unreacted. In addition to the water produced, the fuels C is completely combusted to CO2 and its sulfur content is converted to SO2. In order to evaluate gases emitted at the exhaust stacks for environmental regulation purposes, the nitrogen supplied with the air must also be included in the balanced reactions. a Including the N2 supplied m the air, write a balanced combustion equation for the complex fuel assuming 100% stoichiometric combustion (i.e., when there is no excess oxygen in the products and the only C-containing product is CO2). Except in the case of N2, use only integer coefficients. b Including N2 supplied in the air, write a balanced combustion equation for the complex fuel assuming 120% stoichiometric combustion (i.e., when excess oxygen is present in the products and the only C-containing product is CO2). Except in the case of use only integer coefficients c Calculate the minimum mass (in kg) of air required to completely combust 1700 kg of C11H7S. d Calculate the air/fuel mass ratio, assuming 100% stoichiometric combustion. e Calculate the air/fuel mass ratio, assuming 120% stoichiometric combustion.
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