General, Organic, and Biological Chemistry
General, Organic, and Biological Chemistry
7th Edition
ISBN: 9781285853918
Author: H. Stephen Stoker
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 6, Problem 6.50EP

Balance the following chemical equations.

  1. a. H2S + O2 → SO2 + H2O
  2. b. Ni + HCl → NiCl2 + H2
  3. c. IBr + NH3 → NH4Br + NI3
  4. d. C2H6 + O2 → CO2 + H2O

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The chemical equation H2S+O2SO2+H2O has to be balanced.

Concept Introduction:

Statement that uses chemical symbols and formulas instead of words in order to describe the changes that takes place in a chemical reaction is known as chemical equation.  Reactants are the substances that are present in starting of the chemical reaction and it is entered in left side of the chemical equation.  Products are the substance that is obtained from the reactants as a final outcome of the chemical reaction and it is entered in right side of the chemical equation.  In order to write a chemical equation four conventions are used and they are,

  • Correct formulas of the reactants are written in left side of chemical equation.
  • Correct formulas of products are written in right side of chemical equation.
  • An arrow is placed between reactants and products which point towards the products.
  • If two or more substances are present in reactant and product, plus sign is used to separate them.

Balanced chemical equation is the one that has equal number of atoms of each element that is involved in the chemical reaction on both sides of the chemical equation.  An unbalanced chemical equation can be balanced by adding coefficients to the equation.  It is a number that is placed in left side of the chemical formula so that it changes the amount only and not the identity.

Answer to Problem 6.50EP

The balanced chemical equation is 2H2S+3O22SO2+2H2O.

Explanation of Solution

Given chemical equation is H2S+O2SO2+H2O.  Looking into the equation, the element that is occurring only once on both sides and the compound that has greater number of atoms has to be started with.  Starting with balancing oxygen atom.

Balancing O atoms:  There are two O atoms on left side and three O atoms on the right side of the chemical equation.  Placing the coefficient 3 in front of O2 in the left side, coefficient 2 in front of both H2O and SO2 on right side, balances the O atoms on both sides.  This leaves H and S atoms to become unbalanced.

H2S+3O22SO2+2H2O

Balancing H atoms:  There are two H atoms on left side and four H atoms on the right side of the chemical equation.  Placing the coefficient 2 in front of H2S in the left side balances the H atom on both sides.  This leaves all the atoms on both sides to be balanced.

2H2S+3O22SO2+2H2O

The number of atoms of each element on both sides have to be checked and it can be given as shown below,

Atom Leftside Rightside_H 2x2=4 2x2=4S 2x1=2 1x2=2O 3x2=6 4+2=6_

As there are same number of atoms of each element on both sides of the equation, it is said to be a balanced chemical equation.  The balanced chemical equation can be given as,

2H2S+3O22SO2+2H2O

Conclusion

Given chemical equation is balanced.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The chemical equation Ni+HClNiCl2+H2 has to be balanced.

Concept Introduction:

Statement that uses chemical symbols and formulas instead of words in order to describe the changes that takes place in a chemical reaction is known as chemical equation.  Reactants are the substances that are present in starting of the chemical reaction and it is entered in left side of the chemical equation.  Products are the substance that is obtained from the reactants as a final outcome of the chemical reaction and it is entered in right side of the chemical equation.  In order to write a chemical equation four conventions are used and they are,

  • Correct formulas of the reactants are written in left side of chemical equation.
  • Correct formulas of products are written in right side of chemical equation.
  • An arrow is placed between reactants and products which point towards the products.
  • If two or more substances are present in reactant and product, plus sign is used to separate them.

Balanced chemical equation is the one that has equal number of atoms of each element that is involved in the chemical reaction on both sides of the chemical equation.  An unbalanced chemical equation can be balanced by adding coefficients to the equation.  It is a number that is placed in left side of the chemical formula so that it changes the amount only and not the identity.

Answer to Problem 6.50EP

The balanced chemical equation is Ni+2HClNiCl2+H2.

Explanation of Solution

Given chemical equation is Ni+HClNiCl2+H2.  Looking into the equation, the element that is occurring only once on both sides and the compound that has greater number of atoms has to be started with.  Starting with balancing chlorine atom.

Balancing Cl atoms:  There is one Cl atom on left side and two Cl atoms on the right side of the chemical equation.  Placing the coefficient 2 in front of HCl in the left side, balances the Cl atoms on both sides.  This leaves all the atoms on both sides to be balanced.

Ni+2HClNiCl2+H2

The number of atoms of each element on both sides have to be checked and it can be given as shown below,

Atom Leftside Rightside_Ni 1x1=1 1x1=1H 2x1=2 1x2=2Cl 2x1=2 1x2=2_

As there are same number of atoms of each element on both sides of the equation, it is said to be a balanced chemical equation.  The balanced chemical equation can be given as,

Ni+2HClNiCl2+H2

Conclusion

Given chemical equation is balanced.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The chemical equation IBr+NH3NH4Br+NI3 has to be balanced.

Concept Introduction:

Statement that uses chemical symbols and formulas instead of words in order to describe the changes that takes place in a chemical reaction is known as chemical equation.  Reactants are the substances that are present in starting of the chemical reaction and it is entered in left side of the chemical equation.  Products are the substance that is obtained from the reactants as a final outcome of the chemical reaction and it is entered in right side of the chemical equation.  In order to write a chemical equation four conventions are used and they are,

  • Correct formulas of the reactants are written in left side of chemical equation.
  • Correct formulas of products are written in right side of chemical equation.
  • An arrow is placed between reactants and products which point towards the products.
  • If two or more substances are present in reactant and product, plus sign is used to separate them.

Balanced chemical equation is the one that has equal number of atoms of each element that is involved in the chemical reaction on both sides of the chemical equation.  An unbalanced chemical equation can be balanced by adding coefficients to the equation.  It is a number that is placed in left side of the chemical formula so that it changes the amount only and not the identity.

Answer to Problem 6.50EP

The balanced chemical equation is 3IBr+4NH33NH4Br+NI3.

Explanation of Solution

Given chemical equation is IBr+NH3NH4Br+NI3.  Looking into the equation, the element that is occurring only once on both sides and the compound that has greater number of atoms has to be started with.  Starting with balancing hydrogen atom.

Balancing H atoms:  There are three H atoms on left side and four H atoms on the right side of the chemical equation.  Placing the coefficient 4 in front of NH3 in the left side and coefficient 3 in front of NH4Br, balances the H atoms on both sides.

IBr+4NH33NH4Br+NI3

Balancing Br atoms:  There is one Br atom on left side and three Br atoms on the right side of the chemical equation.  Placing the coefficient 3 in front of IBr in the left side, balances the Br atoms on both sides.  This leaves all the atoms to be balanced on both sides.

3IBr+4NH33NH4Br+NI3

The number of atoms of each element on both sides have to be checked and it can be given as shown below,

Atom Leftside Rightside_N 4x1=4 3+1=4I 3x1=3 1x3=3Br 3x1=3 3x1=3H 4x3=12 3x4=12_

As there are same number of atoms of each element on both sides of the equation, it is said to be a balanced chemical equation.  The balanced chemical equation can be given as,

3IBr+4NH33NH4Br+NI3

Conclusion

Given chemical equation is balanced.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The chemical equation C2H6+O2CO2+H2O has to be balanced.

Concept Introduction:

Statement that uses chemical symbols and formulas instead of words in order to describe the changes that takes place in a chemical reaction is known as chemical equation.  Reactants are the substances that are present in starting of the chemical reaction and it is entered in left side of the chemical equation.  Products are the substance that is obtained from the reactants as a final outcome of the chemical reaction and it is entered in right side of the chemical equation.  In order to write a chemical equation four conventions are used and they are,

  • Correct formulas of the reactants are written in left side of chemical equation.
  • Correct formulas of products are written in right side of chemical equation.
  • An arrow is placed between reactants and products which point towards the products.
  • If two or more substances are present in reactant and product, plus sign is used to separate them.

Balanced chemical equation is the one that has equal number of atoms of each element that is involved in the chemical reaction on both sides of the chemical equation.  An unbalanced chemical equation can be balanced by adding coefficients to the equation.  It is a number that is placed in left side of the chemical formula so that it changes the amount only and not the identity.

Answer to Problem 6.50EP

The balanced chemical equation is 2C2H6+7O24CO2+6H2O.

Explanation of Solution

Given chemical equation is C2H6+O2CO2+H2O.  Looking into the equation, it is found that the carbon-containing compound is oxidized by molecular oxygen to given CO2 and H2O.  For these type of equation, start with balancing hydrogen atoms, followed by carbon atom and then oxygen atom.

Balancing H atoms:  There are six H atoms on left side and two H atoms on the right side of the chemical equation.  Placing the coefficient 6 in front of C2H6 in the left side and coefficient 6 in front of H2O, balances the H atoms on both sides.

2C2H6+O2CO2+6H2O

Balancing C atoms:  There are four C atoms on left side and one C atom on the right side of the chemical equation.  Placing the coefficient 4 in front of CO2 in the right side, balances the C atoms on both sides.

2C2H6+O24CO2+6H2O

Balancing O atoms:  There are two O atoms on left side and fourteen O atoms on the right side of the chemical equation.  Placing the coefficient 7 in front of O2 in the left side, balances the O atoms on both sides.  This leaves all the atoms in both sides to be balanced.

2C2H6+7O24CO2+6H2O

The number of atoms of each element on both sides have to be checked and it can be given as shown below,

Atom Leftside Rightside_C 2x2=4 4x1=4H 2x6=12 6x2=12O 7x2=14 8+6=14_

As there are same number of atoms of each element on both sides of the equation, it is said to be a balanced chemical equation.  The balanced chemical equation can be given as,

2C2H6+7O24CO2+6H2O

Conclusion

Given chemical equation is balanced.

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 6 Solutions

General, Organic, and Biological Chemistry

Ch. 6.3 - Prob. 4QQCh. 6.3 - Prob. 5QQCh. 6.4 - Prob. 1QQCh. 6.4 - Prob. 2QQCh. 6.4 - Prob. 3QQCh. 6.4 - Prob. 4QQCh. 6.5 - Prob. 1QQCh. 6.5 - Prob. 2QQCh. 6.5 - Prob. 3QQCh. 6.5 - Prob. 4QQCh. 6.6 - Prob. 1QQCh. 6.6 - Prob. 2QQCh. 6.6 - Prob. 3QQCh. 6.6 - Prob. 4QQCh. 6.6 - Prob. 5QQCh. 6.7 - Prob. 1QQCh. 6.7 - Prob. 2QQCh. 6.7 - Prob. 3QQCh. 6.7 - Prob. 4QQCh. 6.8 - The problem How many grams of O2 are needed to...Ch. 6.8 - Prob. 2QQCh. 6.8 - How many conversion factors are needed in solving...Ch. 6.8 - Which of the following is the correct conversion...Ch. 6.9 - Prob. 1QQCh. 6.9 - Prob. 2QQCh. 6.9 - Prob. 3QQCh. 6.9 - Prob. 4QQCh. 6 - Calculate, to two decimal places, the formula mass...Ch. 6 - Calculate, to two decimal places, the formula mass...Ch. 6 - The compound 1-propanethiol, which is the eye...Ch. 6 - A compound associated with the odor of garlic on a...Ch. 6 - Indicate the number of objects present in each of...Ch. 6 - Indicate the number of objects present in each of...Ch. 6 - A sample is found to contain 0.500 mole of a...Ch. 6 - A sample is found to contain 0.800 mole of a...Ch. 6 - Select the quantity that contains the greater...Ch. 6 - Select the quantity that contains the greater...Ch. 6 - What is the mass, in grams, of 1.000 mole of each...Ch. 6 - What is the mass, in grams, of 1.000 mole of each...Ch. 6 - How much, in grams, does each of the following...Ch. 6 - How much, in grams, does each of the following...Ch. 6 - How many moles of specified particles are present...Ch. 6 - How many moles of specified particles are present...Ch. 6 - What is the formula mass of a compound whose molar...Ch. 6 - What is the formula mass of a compound whose molar...Ch. 6 - The mass of 7.00 moles of a compound is determined...Ch. 6 - The mass of 5.00 moles of a compound is determined...Ch. 6 - How many moles of oxygen atoms are present in...Ch. 6 - How many moles of nitrogen atoms are present in...Ch. 6 - How many total moles of atoms are present in each...Ch. 6 - Prob. 6.24EPCh. 6 - Write the six mole-to-mole conversion factors that...Ch. 6 - Write the six mole-to-mole conversion factors that...Ch. 6 - Prob. 6.27EPCh. 6 - Based on the chemical formula H2CO3, write the...Ch. 6 - Determine the number of atoms present in 20.0 g...Ch. 6 - Determine the number of atoms present in 30.0 g...Ch. 6 - Determine the mass, in grams, of each of the...Ch. 6 - Determine the mass, in grams, of each of the...Ch. 6 - Determine the number of moles of substance present...Ch. 6 - Determine the number of moles of substance present...Ch. 6 - Determine the number of atoms of sulfur present in...Ch. 6 - Determine the number of atoms of nitrogen present...Ch. 6 - Determine the number of grams of sulfur present in...Ch. 6 - Determine the number of grams of oxygen present in...Ch. 6 - Prob. 6.39EPCh. 6 - Prob. 6.40EPCh. 6 - A compound has a molar mass of 34.02 g. What is...Ch. 6 - A compound has a molar mass of 32.06 g. What is...Ch. 6 - Indicate whether each of the following chemical...Ch. 6 - Indicate whether each of the following chemical...Ch. 6 - Prob. 6.45EPCh. 6 - How many total atoms does each of the following...Ch. 6 - How many oxygen atoms are present on the reactant...Ch. 6 - How many oxygen atoms are present on the reactant...Ch. 6 - Prob. 6.49EPCh. 6 - Balance the following chemical equations. a. H2S +...Ch. 6 - Prob. 6.51EPCh. 6 - Balance the following chemical equations. a. C2H4...Ch. 6 - Prob. 6.53EPCh. 6 - After the following chemical equation was...Ch. 6 - The following diagrams represent the reaction of...Ch. 6 - The following diagrams represent the reaction of...Ch. 6 - Prob. 6.57EPCh. 6 - Prob. 6.58EPCh. 6 - Write the six mole-to-mole conversion factors that...Ch. 6 - Prob. 6.60EPCh. 6 - For the chemical reaction Sb2S3+6HCl2SbCl3+3H2S...Ch. 6 - For the chemical reaction UF6+2H2OUO2F2+4HF write...Ch. 6 - Using each of the following balanced chemical...Ch. 6 - Using each of the following balanced chemical...Ch. 6 - For the chemical reaction C6H12O6+6O26CO2+6H2O how...Ch. 6 - For the chemical reaction C3H8O2+4O23CO2+4H2O how...Ch. 6 - How many water molecules (H2O) are needed to react...Ch. 6 - How many carbon monoxide molecules (CO) are needed...Ch. 6 - The following diagram represents the...Ch. 6 - The following diagram represents the...Ch. 6 - How many moles of beryllium (Be) are needed to...Ch. 6 - How many moles of magnesium (Mg) are needed to...Ch. 6 - The principal constituent of natural gas is...Ch. 6 - Tungsten (W) metal, which is used to make...Ch. 6 - The catalytic converter that is standard equipment...Ch. 6 - A mixture of hydrazine (N2H4) and hydrogen...Ch. 6 - Both water and sulfur dioxide are products from...Ch. 6 - Potassium thiosulfate (K2S2O3) is used to remove...Ch. 6 - How many grams of beryllium (Be) are needed to...Ch. 6 - How many grams of aluminum (Al) are needed to...Ch. 6 - The theoretical yield of product for a particular...Ch. 6 - The theoretical yield of product for a particular...Ch. 6 - Prob. 6.83EPCh. 6 - In an experiment designed to produce calcium oxide...Ch. 6 - If 125.5 g of Ca3N2 were produced from 29.0 g of...Ch. 6 - If 64.15 g of HCl were produced from 2.07 g of H2...
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.
Similar questions
Recommended textbooks for you
Text book image
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY