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High School Chemistry : Balancing Reactions

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Example Questions

Example Question #1 : Balancing Oxidation Reduction 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.

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Example Question #2 : Balancing Oxidation Reduction 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:

One electron is transferred; P is oxidized

Two electrons are transferred; P is oxidized

One electron is transferred; Hg is oxidized

Two electrons are transferred; Hg 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.

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Example Question #2 : Balancing Oxidation Reduction Reactions

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

Possible Answers:

gain . . . gain

gain . . . loss

loss . . . loss

loss . . . gain

None of these; it depends on the reaction.

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.

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Example Question #1 : Balancing Chemical Equations

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

Possible Answers:

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

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

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

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

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

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

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Example Question #1 : Balancing Equations

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

Possible Answers:

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

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

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

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

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

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.

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

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Example Question #2 : Balancing Chemical Equations

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.

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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:

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

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

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Example Question #4 : Balancing Chemical Equations

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

Possible Answers:

$K_3As$

$KAs_3$

$KAs$

$K_2As$

$KAs_2$

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 $K_3As$ .

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Example Question #2 : Balancing Chemical Equations

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

Possible Answers:

$FeSO_4 + CaI_2 \rightarrow FeI_2 +CaSO_4$

$FeSO_4 + CaI \rightarrow FeI +CaSO_4$

None of the available answers

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

$Fe_2SO_4 + CaI_2 \rightarrow 2FeI +CaSO_4$

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:

$FeSO_4 + CaI_2 \rightarrow FeI_2 +CaSO_4$

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Example Question #1 : Balancing Chemical Equations

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:

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

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

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

None of the other choices

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

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.

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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 Oxidation Reduction Reactions

Example Question #2 : Balancing Oxidation Reduction Reactions

Example Question #2 : Balancing Oxidation Reduction Reactions

Example Question #1 : Balancing Chemical Equations

Example Question #1 : Balancing Equations

Example Question #2 : Balancing Chemical Equations

Example Question #4 : Balancing Chemical Equations

Example Question #4 : Balancing Chemical Equations

Example Question #2 : Balancing Chemical Equations

Example Question #1 : Balancing Chemical Equations

All High School Chemistry Resources

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