Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 3.60P

(a)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

_SO2(g)+_O2(g)_SO3(g)

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: 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(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) 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 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

(a)

Expert Solution
Check Mark

Answer to Problem 3.60P

A balanced equation for the following reaction is as follows:

2SO2(g)+O2(g)2SO3(g)

Explanation of Solution

The equation implies that four oxygen (O) atoms are present on the reactant side while three O atoms are present on the product side. Balance O atoms first. To balance O, place the coefficient 2 in front of SO3. The equation becomes,

_SO2(g)+_O2(g)2_SO3(g)

Since one sulfur (S) atom is present on the reactant side while two S atoms are present on the product side. To balance S atoms, place the coefficient 2 in front of SO2. The equation becomes,

2_SO2(g)+_O2(g)2_SO3(g)

Place the coefficient 1 in front of O2 as follows:

2_SO2(g)+1_O2(g)2_SO3(g)

Check whether the equation is balanced or not as follows:

Reactants(2 S, 6 O)Products(2 S, 6 O)

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

A balanced equation for the following reaction is as follows:

2SO2(g)+O2(g)2SO3(g)

(b)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

_Sc2O3(s)+_H2O(l)_Sc(OH)3(s)

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: 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(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) 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. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

(b)

Expert Solution
Check Mark

Answer to Problem 3.60P

The balanced chemical reaction is as follows:

Sc2O3(s)+3H2O(l)2Sc(OH)3(s)

Explanation of Solution

The equation implies that two scandium (Sc) atoms are present on the reactant side while only one Sc atom is present on the product side. Balance Sc atoms first. To balance Sc, place the coefficient 2 in front of Sc(OH)3. The equation becomes,

_Sc2O3(s)+_H2O(l)2_Sc(OH)3(s)

Next, balance hydrogen (H) atoms. Two H atoms are present on the reactant side while six H atoms are present on the product side. To balance H atoms, place the coefficient 3 in front of H2O. The equation becomes,

_Sc2O3(s)+3_H2O(l)2_Sc(OH)3(s)

Hydrogen atoms are balanced. Six oxygen (O) atoms are present on each side of the reaction. Place coefficient 1 in front of Sc2O3.

1_Sc2O3(s)+3_H2O(l)2_Sc(OH)3(s)

Check whether the equation is balanced or not as follows:

Reactants(2 Sc, 6 H, 6 O)Products(1 Sc, 6 H, 6 O)

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

Sc2O3(s)+3H2O(l)2Sc(OH)3(s)

(c)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

_H3PO4(aq)+_NaOH(aq)_Na2HPO4(aq)+_H2O(l)

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: 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(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) 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. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

(c)

Expert Solution
Check Mark

Answer to Problem 3.60P

The balanced chemical reaction is as follows:

H3PO4(aq)+2NaOH(aq)Na2HPO4(aq)+2H2O(l)

Explanation of Solution

Start with Na atoms. The equation implies that one sodium (Na) atom is present on the reactant side while two Na atoms are presents on the product side. To balance Na, place the coefficient 2 in front of NaOH. The equation becomes,

_H3PO4(aq)+2_NaOH(aq)_Na2HPO4(aq)+_H2O(l)

Sodium atoms are balanced, next, balance oxygen (O) atoms. Six O atoms are present on the reactant side while five O atoms are present on the product side. To balance O atoms, place the coefficient 2 in front of H2O. The equation becomes,

_H3PO4(aq)+2_NaOH(aq)_Na2HPO4(aq)+2_H2O(l)

Five hydrogen (H) atoms are present on both sides. Place coefficient 1 in front of H3PO4 and Na2HPO4 as follows:

1_H3PO4(aq)+2_NaOH(aq)1_Na2HPO4(aq)+2_H2O(l)

Check whether the equation is balanced or not as follows:

Reactants(2 Na, 1 P, 5 H, 6 O)Products(2 Na, 1 P, 5 H, 6 O)

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

H3PO4(aq)+2NaOH(aq)Na2HPO4(aq)+2H2O(l)

(d)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

_C6H10O5(s)+_O2(g)_CO2(g)+_H2O(g)

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: 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(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) 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. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

(d)

Expert Solution
Check Mark

Answer to Problem 3.60P

The balanced chemical reaction is as follows:

C6H10O5(s)+6O2(g)6CO2(g)+5H2O(g)

Explanation of Solution

Start with carbon (C) atoms. The equation implies that six C atoms are present on the reactant side while only one C atom is present on the product side. Balance C atoms first and oxygen (O) atoms in last. To balance C, place the coefficient 6 in front of CO2. The equation becomes,

_C6H10O5(s)+_O2(g)6_CO2(g)+_H2O(g)

Carbon atoms are balanced. Next, balance hydrogen (H) atoms. The equation implies that ten H atoms are present on the reactant side while only two H atoms are presents on the product side. To balance H, place the coefficient 5 in front of H2O. The equation becomes,

_C6H10O5(s)+_O2(g)6_CO2(g)+5_H2O(g)

Next, balance oxygen (O) atoms. Seven O atoms are present on the reactant side while seventeen O atoms are present on the product side. Hence, place the coefficient 6 in front of O2. The equation becomes,

_C6H10O5(s)+6_O2(g)6_CO2(g)+5_H2O(g)

Oxygen atoms are balanced. Place coefficient 1 in front of C6H10O5.

1_C6H10O5(s)+6_O2(g)6_CO2(g)+5_H2O(g)

Check whether the equation is balanced or not as follows:

Reactants(6 C, 10 H, 17 O)Products(6 C, 10 H, 17 O)

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

C6H10O5(s)+6O2(g)6CO2(g)+5H2O(g)

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 3 Solutions

Chemistry: The Molecular Nature of Matter and Change

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
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY