Writing and Balancing Chemical Equations

Name ______Date ______Period _____

Writing and Balancing Chemical Equations

Honors and Pre-IBChemistryrefer to chapter 11 in Pearson Chemistry or refer to Chapter 8 in the Modern Chemistry text.

Basics

A chemical reaction is another name for a chemical change and involves the rearranging of atoms by breaking and reforming chemical bonds to produce new compounds with properties different from the original(s). Some signs of a chemical change are: 1) evolution of energy as heat and light. 2) Production of gas. 3) Formation of a precipitant (an insoluble solid, seen as white or colored cloudiness that eventually separates from the solution). 4) Color change.

A chemical equation is a symbolic representation of a chemical reaction containing the following parts:

This equation can be read: “Two moles of hydrogen react with one mole of oxygen to produce 2 moles of water.”

There are 4 parts to a chemical equation. They are the reactant(s), yield sign, product(s), and the coefficients. The reactant(s) come before the yield sign, and make the product(s), which come after the yield sign. The yield sign is in between the reactant(s), and the product(s). It looks like an arrow pointing to the product(s) (). What the yield sign is saying is that the reactant(s), when combined or dissolved, form the product(s). The word yield can be replaced with synonyms such as producing, forming, making, etc.

The following state symbols are often added: s – solid, l – liquid, g – gas, and aq – aqueous (dissolved in water).

Hereis another example:

C10H8(g) + 12O2(g) --> 10CO2(g) + 4H2O(g)

It says: one mole of decene gas plus twelve moles of oxygen gas yields ten moles of carbon dioxide gas and 4 moles of water vapor. We sometimes leave the “moles” out when writing word equations (see balancing below, next page). Hint: If it’s normally a gas at room temperature, say “gas,” if it’s normally a liquid at room temperature, say “vapor.”

Your turn:

1)Label the reactants, products, and coefficients in the following and rewrite it as a word equation (Hint: Use your nomenclature skills for each compound in order to name it):

Na2O + H2O 2NaOH

2) Rewrite the following word equation as a chemical equation and label the reactants, products, and coefficients (Hint: Use your formula writing skills from the prior lessons to come up with the formulas. Do not forget your diatomic elements.):

One mole of solid ammonium dichromate decomposes yielding one mole of nitrogen gas, one mole of solid chromium (III) oxide and 4 moles of water vapor.

Balancing

All chemical equations must obey the Law of Conservation of Matter,“matter is neither gained nor lost during a chemical reaction.” This is the reason coefficients are added. Look at the following equation:

Fe + S8 --> FeS

Note that as it is written above, there is one mole of iron on each side, but there are 8 moles of sulfur in the reactants and only one mole of sulfur in the products. This is not possible according to conservation of matter. We correct this using coefficients. This is known as balancing the equation. Correctly:

8Fe + S8 --> 8FeS

Balancing can be easy or tricky. Either way it takes practice!! Here are some hints:

1)Only add coefficients. NEVER change subscripts. Coefficients can only be placed before each formula, NOT in the middle. They affect everything in that formula (until the plus sign or arrow, but not beyond).

2)Balance the different types of atoms one at a time. Often balancing one atom may unbalance others. That’s ok. Just keep working it!

3)First balance the atoms of elements that are combined and that appear only once on each side of the equation.

4)Balance polyatomic ions that appear on both sides of the equation as single units.

5)Balance H and O atoms and any others that repeat often or are single (by themselves) after atoms of all other elements have been balanced.

Note: It may be helpful, especially at first, to make a chart using E for elements, R for reactants, and P for products in order to help you keep count.

Example 1- Balance the following reaction and rewrite it as a word equation:

Solid zinc plus aqueous hydrochloric acid forms aqueous zinc chloride and hydrogen gas.

The equation: Zn(s) + HCl(aq) ZnCl2(aq) + H2(g)

Strike through's are numbers I changed after adding a coefficient. My chart and final answer showing my work:

E / R / P

Zn11

H1 22

Cl1 22

Step 1: Note that all the numbers in R column do not match the numbers in the P column, so you need to balance. H is left until last. I choose to start with Cl by adding a 2 in front of HCl in the reactants. Recounting everything shows I have balanced the H as well and I’m done!

The answer: Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Notice that all my R and P numbers for each element are the same when I’m through.

The word equation is: Solid zinc reacts with hydrochloric acid forming aqueous zinc chloride and hydrogen gas.

Your turn:

Balance the following.

C(s) + O2(g) CO2(g)

Example 2

An aqueous solution of aluminum sulfate is mixed with an aqueous solution of calcium hydroxide forming aqueous aluminum hydroxide and a solid calcium sulfate precipitant.

The equation: Al2(SO4)3(aq) + Ca(OH)2(aq) Al(OH)3(aq) + CaSO4(s)

Strike through's are numbers I changed after adding a coefficient. My chart and final answer showing my work:

E / R / P

Al 21 2

SO4 3 1 3

Ca 1 3 1 3

OH 2 6 3 6

Step 1: Calcium appears balanced at the beginning. I can begin with either OH, SO4, or Al. I chose aluminum because it looks simpler and add a 2 in front of Al(OH)3 (this also affects OH) and recount my chart. It is good that I got an even number, 6, for OH, because the OH in the R column is a 2 which means I must end up with an even number of OH in my final equation.

Step 2: Next, I chose to work on OH by adding a 3 in front of Ca(OH)2 in the reactants. Again, I recount. I note that my Ca and SO4 are now both off by a factor of 3 and they are together in the products, what fortune!!

Step 3: I add a 3 in front of CaSO4 in the products and recounting find I have balanced my equation!!!

Step 1: Al2(SO4)3(aq) + Ca(OH)2(aq) 2Al(OH)3(aq) + CaSO4(s)

Step 2: Al2(SO4)3(aq) + 3Ca(OH)2(aq) 2Al(OH)3(aq) + CaSO4(s)

Step 3: Al2(SO4)3(aq) + 3Ca(OH)2(aq) 2Al(OH)3(aq) + 3CaSO4(s) answer

(Note: I have rewritten the equation 3 times for clarity in my examples. You can just add the coefficients as you go without rewriting. You can even strike through a coefficient and write the new one above it if you need to change it. Just rewrite the final answer if your work gets too messy.)

Your turn:

Balance the following and rewrite it as a word equation.

Al(s) + Pb(NO3)2(aq)  Pb(s) + Al(NO3)3(aq)

Write and balance the following equation.

An aqueous solution of ammonium sulfate is combined with an aqueous solution of silver nitrate forming aqueous ammonium nitrate and a solid silver sulfate precipitate.

Now for a more difficult example!

Example3-Write and balance the following reaction:

Solid aluminum carbide reacts with liquid water forming methane gas and solid aluminum hydroxide.

First, write the equation: Al4C3(s) + H2O(l) CH4(g) + Al(OH)3(s) Hint: Leave room to write in coefficients.

Next, list the elements in your chart and count up the reactant and product amounts.

For the equation above:

E / R / P

Al41

C31

H23

O13

Step 1: Note that all the numbers in R column do not match the numbers in the P column, so you need to balance. Also, note that hydrogen repeats and its best to leave H and O until last. You may start with C or Al. I chose aluminum, but this will affect my H and O count too! I will add a 4 in front of Al(OH)3in order to balance the Al. Now, recount everything affected.

Step 2: Next, balance the carbon by adding a 3 in front of methane, but this affects the H count as well, so change it in your chart again.

Step 3: Since oxygen does not repeat like hydrogen, I work on O next by adding a 12 in front of H2O. Time to recount and update my chart…and hydrogen balances itself! This may not always happen, but often it will. Since my R and P columns match, I’m done and the third equation below is the answer .

Step 1: Al4C3(s) + H2O(l) CH4(g) + 4Al(OH)3(s)

Step 2: Al4C3(s) + H2O(l) 3CH4(g) + 4Al(OH)3(s)

Step 3: Al4C3(s) + 12H2O(l) 3CH4(g) + 4Al(OH)3(s) answer

E / R / P

Al41 4

C31 3

H2 247 16 24

O1 123 12

Your turn: Balance the following equationsand also rewrite them as word equations. Show your work.

1) H2O(g)  H2(g) + O2(g)

2) Mg + HCl  MgCl2 + H2

3) Mg(s) + O2  MgO(s)

4) AgNO3 + KBr  AgBr + KNO3

5) P4(s) + O2  P2O5(s)

Write and balance the following equations.

6)Aqueous aluminum sulfate reacts with aqueous barium chloride yielding aqueous aluminum chloride and solid barium sulfate.

7)Solid copper metal reacts with aqueous silver nitrate producing solid silver metal and aqueous copper (II) nitrate.

8)Aqueous sodium hydroxide plus aqueous iron (III) chloride forms aqueous sodium chloride and an iron (III) hydroxide precipitant (recall that precipitants are solids).

9)Aqueous potassium chloride reacts with aqueous lead (II) nitrate yielding aqueous potassium nitrate and a lead (II) chloride precipitant.

10)Solid cerium (IV) oxide and solid potassium iodide are added to an aqueous hydrochloric acid solution forming aqueous potassium chloride, aqueous cerium (III) chloride, liquid water, and solid iodine.

Chemical Reactions Involve Energy Changes

All chemical changes involve energy. Bond breaking is endothermic (uses energy). Bond making creates a more stable, lower energy state and is exothermic (releases energy). The net enthalpy of a reaction, ΔH, (approximately equal to heat of the reaction) depends on if the reactants use more energy or the products release more energy.

The following graphs show the energy relationships of reactions:

The above graphs appear commonly on exams. Know them well!!!

Note: ΔH of an exothermic reaction is negative (-ΔH = energy released) and ΔH of an endothermic reaction is positive (+ΔH = energy stored). Ea is the activation energy needed to start the reaction.

Another important concept common on tests is the effect of a catalyst on the graphs:

Note: Acatalyst speeds up the reaction by lowering the activation energy.

Finally, reaction enthalpy can be expressed as a mole ratio. For example, look at the equation for cellular respiration:

C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l) + 2870 kJ ΔH = -2870 kJ/mol C6H12O6

This tells us that each mole of glucose produces 2870 kJ of energy, that is 2870 kJ/mol for the reaction. This number can be used as a conversion factor. Since the enthalpy is on the products side, the reaction is exothermic. If the enthalpy is a reactant, the reaction is endothermic.

CaCO3(s) + 178 kJ  CaO(s) + CO2(g) ΔH = +178 kJ/mol CaCO3

Example problem: How much energy is needed to decompose 26.3 g CaCO3?

Plan: Change g to mol and mol to kJ

Step 1: Find the molar mass of CaCO3. 40.08 + 12.01 + (3)(16.00) = 108.09 g/mol CaCO3

Step 2: Set up the cross starting with the given over one and converting g to mol, then mol to kJ. Calculate and round to 3 significant figures.

(26.3 g CaCO3/1)(1 molCaCO3/108.09 gCaCO3)(+178 kJ/mol CaCO3) = 43.3 kJ

Your turn:

Answer the next several questions using the graphs below.

1. Which letter on the graphs represents the product of the chemical reaction?
2. Which letter on the graphs represents the activation energy?
3. Which letter on the graphs represents energy released from the reactants?
4. Which letter on the graphs represents the reactants of the chemical reaction?
5. Which letter on the graphs represents energy gained or stored in the products?
6. Which graph represents an endothermic reaction?
7. Which graph represents an exothermic reaction?
8. Draw graph I over showing how a catalyst will affect the graph.

1. The formation of ammonia involves the following reaction, known as the Haber Process:

N2(g) + H2(g)  NH3(g) + 92.0kJ Remember ΔH = -92.0 kJ/mol of reaction.

Hint: Balancetheequation before you go any further!!!

a)How many kilojoules of energy are involved in the above reaction if 120 g of nitrogen gas react with excess hydrogen gas?

b)Is the reaction exothermic or endothermic? Explain.

Types of Reactions

Inorganic reactions fall into a number of categories, although some reactions may fit in more than one category. The main reaction types are:

1)Synthesis

2)Decomposition

3)Single replacement (also known as single displacement)

4)Double replacement (also known as double displacement)

5)Combustion

6)Oxidation-reduction

I hope to demonstrate these reaction types to you soon in class.

Synthesis (Also known as “combination”)

Synthesis involves the combining of reactants to form a single product in the form: A + B  AB

Synthesis example: N2(g) + 3H2(g)  2NH3(g)

Decomposition

Decompositionis essentially the opposite of synthesis andinvolves the splitting apart of a single reactant into two or more product in the form: AB  A + B

Decomposition example: 2KClO3(s)  2KCl(s) + 3O2(g)

Single Replacement

In a single replacement, a more reactive compound replaces a less reactive compound in the form:

A + BC  AC + B

Single replacement example: 2Al(s) + 3CuCl2(aq) 2AlCl3(aq) + 3Cu(s) Note: the more active metal aluminum replaces the less active copper.

Whether or not a single replacement reaction will occur can be predicted using a activity series that was experimentally derived. A more active substance will replace a less active substance.

On most charts an element will be replaced by one above it, but not below.

Generally, the further apart the elements are, the more likely the replacement will occur.

Your Turn

1)Using the activity table above, list the following elements from least to most reactive:

Zn, Cu, Mg, Au, K, Ca, Sn

2) Predict the product. If the reaction is not likely to occur, put NR (No Reaction) after the arrow. If the reaction does occur, write a balanced chemical equation for the reaction.

a)Chlorine gas and aqueous potassium iodide.

b)Magnesium metal and aqueous copper (II) sulfate.

c)Copper metal and aqueous iron (III) chloride.

d)Lead metal and aqueous silver nitrate.

Double Replacement

Double replacement reactions involve reactions in which elements “swap partners” in the form:

AB + CD  AD + CB

Often a double replacement involves the mixing of two clear solutions resulting in the formation of an insoluble solid known as a precipitant.

Double replacement example: Pb(NO3)2(aq) + 2KI(aq)  2KNO3(aq) + PbI2(s) Note: lead (II) iodide is the precipitant.

It is usually more useful to write double replacement reactions as net ionic equations. A net ionic equation only shows the ions that actually react. The ions that do not react are called spectator ions.

Example

For the equation: Pb(NO3)2(aq) + 2KI(aq)  2KNO3(aq) + PbI2(s)

Step 1: First, we write an ionic equation by by showing all the ions separately. Remember that only aqueous compounds are separated into ions, not solids, liquids, nor gases.

Word equation: Aqueous lead (II) nitrate and potassium iodide are mixed forming solid lead (II) iodide and aqueous

potassium nitrate.

Ionic equation: Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  2K+(aq) + 2NO3-(aq) + PbI2(s)

Step 2: Next, we cross off the spectator ions, that is those found the same on both sides of the arrow.

Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  2K+(aq) + 2 NO3(aq) + PbI2(s)

Step 3: Finally, we rewrite the remaining ions and compounds as a simplifiied net ionic equation.

Pb+2(aq) + 2I-(aq)  PbI2(s)

Your Turn Write a balanced chemical equation, ionic equation, and net ionic equation for each of the following. Include state symbols (s, l, g, aq) and identify any precipitants.

1) Aqueous solutions of lead (II) nitrate and ammonium chloride are mixed, forming solid lead (II) chloride and aqueous ammonium nitrate.

2)Aqueous ammonium chloride and sodium carbonate are combined, producing solid ammonium carbonate and another aqueous product you are to determine.

Combustion

Combustion is an exothermic reaction in which an element combines with oxygen gas (O2). Combustion gives off heat!

Example: 2Mg(s) + O2(g)  2MgO(s)

A common type of combustion is complete combustion of a hydrocarbon. This reaction is the reaction by which most of our fuels burn. Hydrocarbons are compounds containing carbon chains surrounded by only covalently bonded hydrogen. Examples are methane, propane, and butane that are used around the home as fuels. The reactants of this reaction are always just hydrocarbon and oxygen gas. The products of complete combustion of a hydrocarbon are always just carbon dioxide and water. Substituted hydrocarbons, like alcohols burn in a similar reaction.

Example: Methane burns completely. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

Oxidation-Reduction

Oxidation-Reduction reactions are also known as redox reactions and may also be in several of the above categories as well as being redox.

Oxidation means to lose electrons and reduction means to gain electrons. We remember OIL RIG. Oxidation Is Lose. Reduction Is Gain.

Oxidation-Reduction can be complicated and involves calculating oxidation states and balancing half-reactions showing only oxidation or reduction separately. We may look at these in more detail later, depending on time. If you ever take a more advanced course you will definitely learn about these reactions.

Example: I will not show you here how I figure out the electrons movements because it is somewhat involved.

From the single replacement section above: 2Al(s) + 3CuCl2(aq) 2AlCl3(aq) + 3Cu(s)

In this reaction, aluminum transfers 3 electrons to copper forming an aluminum ion in aluminum chloride and precipitating copper metal from the copper (II) ions in copper (II) chloride. The copper is seen as a red, flakey or clumpy precipitant. The unbalanced oxidation half reaction is Al Al+3 + 3e- and the unbalanced reduction half reaction is Cu+2 + 2e- Cu. Balancing the half reactions so the electron amounts are equal requires considering whether it occurs in acid or base.

Your Turn

Rewrite each word equation as abalanced chemical equation showing as many reaction conditions and other information as possible. Also identify the reaction type for each. For any double replacement reactions, write a net ionic equation.

1. Silver nitrate reacts with calcium chloride in aqueous solution to form calcium nitrate and a silver chloride precipitant.

1. Hydrogen gas reacts with chlorine gas and is bubble through water to form aqueous hydrochloric acid.

1. Propane, C3H8, undergoes complete combustion.

1. Metallic sodium reacts to replace the hydrogen in aqueous hydrochloric acid.

1. Solid calcium carbonate breaks down when heated to form solid calcium oxide and carbon dioxide gas.

Rewrite each chemical equation below as a word equation with all possible information and identify the reaction type. Also balance the equations when necessary.

1. Al(s) + O2(g) Al2O3(s)

1. CS2(g) + O2(g) SO2(g) + CO2(g) Hint: The reaction type is tricky here. Think of the O2 as O-O.

1. MgI2(aq) + Br2(g) MgBr2(aq) + I2(g)

Handout by: M. Silverman 1