We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

High School Chemistry : Balancing Reactions

Study concepts, example questions & explanations for High School Chemistry

Create An Account

All High School Chemistry Resources

6 Diagnostic Tests 143 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Balancing Reactions

What is the coefficient on sulfur dioxide if the following redox reaction is balanced in an acidic solution?

$SO_{2} + Cr_{2}O_{7}^{2-} \rightarrow SO_{4}^{2-} + Cr^{3+}$

Possible Answers:

Correct answer:

Explanation:

Balancing redox reactions involves the following steps:

1. Divide the reaction into oxidation and reduction half reactions and balance all atoms that are not oxygen and hydrogen:

$SO_{2} \rightarrow SO_{4}^{2-}$

$Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+}$

2. Balance the oxygens by adding water molecules on the opposite side of the reactions:

$2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-}$

$Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

3. Balance the hydrogens by adding protons to the opposite side of the equation:

$2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-} + 4H^{+}$

$14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

4. Add electrons in order to equal the charges on both sides of the equation:

$2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-} + 4H^{+} + 2e^{-}$

$6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

5. In order to make the electron exchange equal in each half step, we must multiply the top half reaction by 3:

$6H_{2}O + 3SO_{2} \rightarrow 3SO_{4}^{2-} + 12H^{+} + 6e^{-}$

$6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

6. Add up the reactants and products while cancelling out substances on opposite sides of the reactions. For example: we will cancel the 6 water molecules as reactants and be left with only one water molecule as a product. In addition, only 2 protons will be left on the reactant's side after canceling the 12 from the product's side.

$Cr_{2}O_{7}^{2-} + 3SO_{2} + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_{4}^{2-} + H_{2}O$

In the balanced redox reaction, the coefficient on sulfur dioxide is 3.

Report an Error

Example Question #31 : Chemical Reactions

For the following unbalanced redox reaction, how many electrons are transferred and which chemical species is being oxidized?

$PF_2I_{(l)}+Hg_{(l)}\rightarrow P_2F_4_{(g)}+Hg_2I_2_{(s)}$

Possible Answers:

Two electrons are transferred; Hg is oxidized

One electron is transferred; Hg is oxidized

Two electrons are transferred; P is oxidized

One electron is transferred; P is oxidized

Correct answer:

Two electrons are transferred; Hg is oxidized

Explanation:

To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of . Phosphorus begins with an oxidation number of and ends with an oxidation number of . Note that the oxidation numbers for fluorine and iodine reamain constant at for each.

Now we can begin to look at the half-reactions.

$PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Balance the atoms.

$2PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$2Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.

$2PF_2I_{(l)}+2e^-\rightarrow P_2F_4_{(g)}$

$2Hg \rightarrow Hg_2I_2_{(s)}+2e^{-}$

We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.

Report an Error

Example Question #241 : High School Chemistry

Oxidation is the __________ of electrons, reduction is the __________ of electrons.

Possible Answers:

None of these; it depends on the reaction.

loss . . . gain

gain . . . gain

loss . . . loss

gain . . . loss

Correct answer:

loss . . . gain

Explanation:

An oxidation-reduction (redox) reaction is a reaction where electrons are transferred between two substances. When an atom is oxidized, it is called the reducing agent, and it loses a number of electrons. Similarly, a reduced atom is called the oxidizing agent, and gains the same number of electrons. A popular mnemonic to help remember is OIL RIG, or Oxidation is Loss of electrons, Reduction is Gain of electrons.

Report an Error

Example Question #71 : Solutions

What is the formula for the dissociation of iron (II) phosphate?

Possible Answers:

$FeP\rightleftharpoons Fe^{2+}+P^{2-}$

$Fe_2(PO_4)_3\rightleftharpoons 2Fe^{3+}+3PO_4^{2-}$

$FePO_4\rightleftharpoons Fe^{2+}+PO_4^{2-}$

$Fe_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$

$Ir_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$

Correct answer:

$Fe_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$

Explanation:

Iron (II) has a positive two charge: $Fe^{2+}$ .

Phosphate has a negative three charge: $PO_4^{3-}$ .

The initial compound must be constructed to cancel these charges. The dissociation is: $Fe_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$ .

Report an Error

Example Question #32 : Chemical Reactions

Calcium hydroxide is treated with hydrochloric acid to produce water and calcium chloride. Write a balanced chemical reaction that describes this process.

Possible Answers:

$2Ca(OH)_{2} + HCl \rightarrow H_{2}O + 2CaCl_{2}$

$Ca(OH)_{2} + HCl \rightarrow H_{2}O + CaCl_{2}$

$Ca(OH)_{3} + 3HCl \rightarrow 3H_{2}O + CaCl_{3}$

$CaOH + HCl \rightarrow H_{2}O + CaCl$

$Ca(OH)_{2} + 2HCl \rightarrow 2H_{2}O + CaCl_{2}$

Correct answer:

$Ca(OH)_{2} + 2HCl \rightarrow 2H_{2}O + CaCl_{2}$

Explanation:

Calcium is in the second group of the periodic table, and is therefore going to have a $\small +2$ oxidation number. Hydroxide ions have a $\small -1$ charge. Calcium hydroxide will have the formula $\small Ca(OH)_2$ .

Chloride ions have a $\small -1$ charge and hydrogen ions have a $\small +1$ charge. The formula for hydrochloric acid is $\small HCl$ .

On the products side, water has the formula $\small H_2O$ and calcium chloride has the formula $\small CaCl_2$ .

Now that we know all of the formulas, we can write our reaction:

$Ca(OH)_2+HCl\rightarrow H_2O+CaCl_2$

In order to balance the chloride atoms, we need to add coefficients.

$Ca(OH)_2+2HCl\rightarrow 2H_2O+CaCl_2$

Report an Error

Example Question #1 : Balancing Reactions

What is the coefficient for oxygen gas when the following equation is balanced?

$C_{5}H_{12} + O_{2} \rightarrow CO_{2} + H_{2}O$

Possible Answers:

Correct answer:

Explanation:

The balanced reaction for the combustion of pentane is:

$C_{5}H_{12} + 8O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

When balanced, oxygen gas has a coefficient of eight.

To balance the equation, it is easiest to leave oxygen and hydrogen for last. This means we should start with carbon.

$C_{5}H_{12} + O_{2} \rightarrow CO_{2} + H_{2}O$

$C_{5}H_{12} + O_{2} \rightarrow 5CO_{2} + H_{2}O$

Now that carbon is balanced, we can look at hydrogen.

$C_{5}H_{12} + O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

Finally, we can balance the oxygen.

$C_{5}H_{12} + 8O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

The final reaction uses five carbon atoms, twelve hydrogen atoms, and sixteen oxygen atoms per side.

Report an Error

Example Question #4 : Balancing Chemical Equations

Given the unbalanced reaction below, how many moles of solid iron can be made from ten moles of iron oxide?

$Al_(_s_) + Fe_2O_3_(_s_) \rightarrow Al_2O_3_(_s_) + Fe_(_s_)$

Possible Answers:

10 mol

20 mol

1 mol

5 mol

2 mol

Correct answer:

20 mol

Explanation:

The balanced chemical equation is:

$\tiny \small \small 2Al_(_s_) + Fe_2O_3_(_s_) \rightarrow Al_2O_3_(_s_) + 2Fe_(_s_)$

The mole ratio of iron oxide to solid iron is 1:2. You can set up the following proportion to solve:

$\frac{1mol\ Fe_2O_3}{2mol\ Fe}=\frac{10mol\ Fe_2O_3}{x}$

$x = (2)(10) = 20 mol\ Fe$

Report an Error

Example Question #2 : Introduction For Reactions

What is the chemical formula of the salt formed when a chemist mixes solvated Potassium and Arsenic ions in solution?

Possible Answers:

$K_3As$

$KAs$

$KAs_3$

$K_2As$

$KAs_2$

Correct answer:

$K_3As$

Explanation:

Potassium is a Group I element, so to get to a filled valence shell, it will lost one electron, yielding $K^+$ .

Arsenic is a Group 5 element, so it needs to gain three electrons to obtain a filled valence shell, yielding $As^{-3}$ .

In order to balance out the charges, the resultant salt will be $K_3As$ .

Report an Error

Example Question #2 : Balancing Reactions

What is the net ionic equation for the ion exchange reaction between ferrous sulfate and calcium iodide? Assume all compounds are soluble.

Possible Answers:

$FeSO_4 + CaI \rightarrow FeI +CaSO_4$

None of the available answers

$Fe(SO_4)_2 + CaI_2 \rightarrow FeI_2 +Ca(SO_4)_2$

$Fe_2SO_4 + CaI_2 \rightarrow 2FeI +CaSO_4$

$FeSO_4 + CaI_2 \rightarrow FeI_2 +CaSO_4$

Correct answer:

$FeSO_4 + CaI_2 \rightarrow FeI_2 +CaSO_4$

Explanation:

First, we must know what ferrous sulfate is. Ferrous refers to $Fe^{2+}$ , and sulfate has the formula $SO_4^{-2}$ . When we combine the two together we get FeSO_4 .

Calcium is a divatent cation and iodide is a monovalent anion, so their salt is CaI_2 . The ion exchange reaction is then:

$FeSO_4 + CaI_2 \rightarrow FeI_2 +CaSO_4$

Report an Error

Example Question #3 : Introduction For Reactions

Select the net ionic equation from this molecular reaction:

$Ba(NO_{3})_{2} (aq) + K_{2}SO_{4} (aq) \rightarrow 2KNO_{3} (aq) + BaSO_{4} (s)$

Possible Answers:

$K^{+} (aq) + NO_{3}^{-} (aq) \rightarrow KNO_{3} (aq)$

$Ba^{2+} (aq) + 2K^{+} (aq) + SO_{4}^{2-} (aq) \rightarrow 2K^{+} (aq) + BaSO_{4} (s)$

$Ba^{2+} (aq) + SO_{4}^{2-} (aq) \rightarrow BaSO_{4} (s)$

None of the other choices

$2K^{+} (aq)+SO_{4}^{2-}(aq)\rightarrow K_{2}SO_{4} (aq)$

Correct answer:

$Ba^{2+} (aq) + SO_{4}^{2-} (aq) \rightarrow BaSO_{4} (s)$

Explanation:

The net ionic equation is derived by removing all spectator ions from the total ionic equation (in which all ions are listed). To put it another way, the net ionic equation involves only the ions that participate in a reaction which, in this case, is the precipitation of barium sulfate.

Begin by writing all aqueous compounds in their dissociated (ionic) forms.

$Ba(NO_{3})_{2} (aq) + K_{2}SO_{4} (aq) \rightarrow 2KNO_{3} (aq) + BaSO_{4} (s)$

$Ba^{2+}+2NO_3^-+2K^++SO_4^{2-}\rightarrow2K^++2NO_3^-+BaSO_4(s)$

Cancel out any ions that appear in equal quantities on both sides of the equation. In this case, we can cancel the nitrate and potassium ions.

$Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s)$

This is our net ionic equation.

Report an Error

View Tutors

Andrew
Certified Tutor

The University of Kansas, Bachelor of Science, Biochemistry.

View Tutors

Mr. Henderson
Certified Tutor

Swarthmore College, Bachelor of Science, Engineering. MBU, Master of Science, Education.

View Tutors

Justin
Certified Tutor

University of Toledo, Bachelors, Environmental Science. University of Michigan, Masters, Natural Resource Management.

All High School Chemistry Resources

6 Diagnostic Tests 143 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

SSAT Tutors in Los Angeles, Statistics Tutors in Seattle, Spanish Tutors in Chicago, SSAT Tutors in Houston, SAT Tutors in Boston, SSAT Tutors in Denver, Physics Tutors in Boston, SAT Tutors in New York City, SAT Tutors in Dallas Fort Worth, Math Tutors in Phoenix

Popular Courses & Classes

SSAT Courses & Classes in Seattle, MCAT Courses & Classes in Houston, LSAT Courses & Classes in Phoenix, GMAT Courses & Classes in Los Angeles, Spanish Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in Houston, GMAT Courses & Classes in Philadelphia, SAT Courses & Classes in Los Angeles, LSAT Courses & Classes in New York City, Spanish Courses & Classes in New York City

Popular Test Prep

MCAT Test Prep in Washington DC, GMAT Test Prep in Dallas Fort Worth, GRE Test Prep in Chicago, SSAT Test Prep in Seattle, GMAT Test Prep in Denver, MCAT Test Prep in Miami, GRE Test Prep in Houston, SAT Test Prep in Philadelphia, GRE Test Prep in San Diego, LSAT Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

High School Chemistry : Balancing Reactions

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #1 : Balancing Reactions

Example Question #31 : Chemical Reactions

Example Question #241 : High School Chemistry

Example Question #71 : Solutions

Example Question #32 : Chemical Reactions

Example Question #1 : Balancing Reactions

Example Question #4 : Balancing Chemical Equations

Example Question #2 : Introduction For Reactions

Example Question #2 : Balancing Reactions

Example Question #3 : Introduction For Reactions

All High School Chemistry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: