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Balancing Chemical Equations

Things to remember

1. There must be the same number of atoms of each element on both side of the yield sign(-->) A numerical subscript next an element indicates how many atoms of that element there are in that molecule. For example, in H2O the subscript 2 indicates that there are two hydrogen atoms. There is no subscript on the oxygen thus indicating that there is only one oxygen atom in this water molecule.
2.

3. When we put a number in front of a molecule this indicates the number of molecules present. For example, 2 H2O indicates that there are 2 water molecules. This changes the number of atoms present, now there are 4 hydrogen atoms and 2 oxygens.
4.

5. If you are having difficulties getting an equation to balance go back and check all the chemical formula to make sure that they are correct.
6. Remember the rule of balancing.

Tips

2. If a polyatomic appears on both sides of the equation you can treat it as if it were a single element or what Mrs. Warnick likes to call a family.
3. In a combustion reaction leave the oxygen until last.

Examples

2. Mg + HNO3 --> Mg(NO3)2 + H2

Step 1. Start with Mg. There is 1 Mg on the reactant side and 1 on the product side so nothing needs to be done with Mg at this point.

Step 2. NO3 appears on both sides of the equation so it can be treated as a family. There is one NO3on the reactant side and 2 on the product [ (NO3)2]. The least common multiple is 2 so each side need to have a total of 2 NO3on each side. This is accomplished by putting a 2 in front of the HNO3 on the reactant side.

Step 3. Now look at hydrogen. On the reactant side there are 2 atoms and there are two atoms on the products side. When you put a 2 in front of the HNO3 this changed the number of hydrogen atoms on the reactant side.

Answer: Mg + 2HNO3 --> Mg(NO3)2 + H2

4. C6H6 + O2 --> CO2 + H2O

Step 1. Start with the carbon. There are six on the reactant side and only one on the product side.

C6 -->C

The least common multiple is 6. So there must be 6 C on both sides of the arrow. In order to do this you must multiple the product C by 6.

C6 -->6C

Step 2. Now balance the hydrogen. There are 6 H on the reactant side and only 2 on the product side.

H6 --> H2

The least common multiple is 6. So there must be 6 H on both sides of the arrow. In order to do this you must multiple the product H by 3.

H6 -->3 H2

Step 3. Now balance the oxygen. There are 2 O on the reactant side and 15 on the product side.

O2 -->6CO2 + 3H2O

The least common multiple is 30. So there must be 30 O on both sides of the arrow. In order to do this you must multiple the reactant O by 15. Then you need to multiple the product Os by 2.

15O2 --> 12CO2 + 6H2O

Remember that if there are more than one compound with an element on the same side of the equation all of these compounds must be multiplied.

Step 4. Because you changed the number in front of compounds that contained elements that we have already balanced you must go back and make sure that these are balanced again.

Carbon: 6 -->12.

Least common multiple is 12. Multiple the reactant by 2.

2C6H6 + 15 O2 -->12CO2 + 6H2O

Hydrogen: 12 -->12.

Answer : 2C6H6 + 15 O2 -->12CO2 + 6H2O

5. Ag + S -->Ag2S

Step 1. Start with the silver. On the reactant side there a one silver and on the product side there are two silvers.

Ag --> Ag2S

The least common multiple is 2. So there must be two Ag on both sides of the arrow. In order to do this you must multiple the reactant Ag by 2.

2Ag --> Ag2S

Step 2. Look at the sulfur. There is a total of one sulfur on each side so it is balanced.

Answer: 2Ag + S --> Ag2S

9. CaO + P2O5 --> Ca3(PO4)2

Step 1. Start with the calcium. There is 1 on the reactant side and 3 on the product side.

Ca --> Ca3(PO4)2

The least common multiple is 3. So there needs to be a total of 3 Ca on both sides of the arrow. This can be done by multiplying the calcium on the reactant side by 3.

3 Ca --> Ca3(PO4)2

Step 2. Now balance the oxygen. (You can not balance PO4 as a family because it does not appear on both sides of the arrow.) There are 8 oxygen on the reactant side and 8 on the product side.

3CaO + P2O5 --> Ca3(PO4)2

Nothing needs to change.

Step 3. Now balance the phosphorous. There are 2 P on the reactant side and 2 P on the product side.

3CaO + Ca3(PO4)2 --> Ca3(PO4)2

Nothing needs to change.

Answer: 3CaO + Ca3(PO4)2 --> Ca3(PO4)2

The same procedures are followed for all balancing problems. Check yourself to make sure you are doing them right. If you are still having troubles. Talk with your teacher before or after school.

On the following remember that if you can not get it to balance check your formulas.

12. 2NaOH + H2SO4 -->Na2SO4 + 2HOH

14. 2C4H10 + 13O2 -->8CO2 + 10H2O

18. ZnCl2 + (NH4)2S -->ZnS + 2NH4Cl

23. Mn(ClO4)4 -->MnCl4 + 8O2

25.2Na + 2HOH -->2NaOH + H2

29. Pb(C2H3O2) + 2Ag2S -->PbS2 + 4Ag(C2H3O2)

33. H2 + Cl2 -->2HCl

35. 3Cl2 + 2AuI3 -->2AuCl3 + 3I2

Identifying Types of Reactions

Tips

Use your packet as a reference until you get familiar with identifying the types of reactions.

Examples

2. Mg + HNO3 --> Mg(NO3)2 + H2

Step 1. Identify the reactants. This reaction has two reactants which rules out a decomposition. One of the reactants is an element and the other is a compound.

Step 2. Identify the products. This reaction produces two products. One of the products is an element and the other is a compound.

Step 3. From steps 1 and 2 identify the type of reaction. Because both sides have an element and a compound the only logical type of reaction would be a single replacement. If you look at the equation you can see that it is a single replacement because Mg (a metal) replaces hydrogen.

4. C6H6 + O2 --> CO2 + H2O

Step 1. Identify the reactants. This reaction starts off with a carbon-hydrogen compound and oxygen.

Step 2. Identify the products. The products produced are carbon dioxide and water.

Step 3. Identify the type of reaction. This could be a double replacement except for the fact that there is no replacement taking place. It may look like oxygen is replacing hydrogen, but this is not true because both products contain oxygen. Because the reactants are carbon-hydrogen compound and the oxygen this has to be a combustion reaction.

5. Ag + S -->Ag2S

Step 1. Identify the reactants. This reaction has two reactants and both are single elements.

Step 2. Identify products. There is only one product and it is a compound made up of the two element reactants.

Step 3. Identify the type of reaction. Because two single element reactants combine to form one compound product this must be a composition reaction.

9. CaO + P2O5 --> Ca3(PO4)2

Step 1. Identify the reactants. There are two compounds as reactants.

Step 2. Identify the products. There is only one product and it is also a compound.

Step 3. Identify the type of reaction. Because there are two reactants combining to form only one product this must be a composition reaction. Composition reactions do not always have to be the combining of two elements to form a compound.

The same procedures are followed for all balancing problems. Check yourself to make sure you are doing them right. If you are still having troubles. Talk with your teacher before or after school.

12. Double replacement.

14. Combustion.

18. Double replacement

23. Decomposition

25.Single replacement

29. Double replacement

33. Composition

35. Single replacement

Stoichiometry

Stoichiometry is the language and math used to determine and communicate the amount of reactants and products involved in a chemical reaction.

Things to Remember

1. Make sure that your chemical equation is correctly balanced.
2. Always decide what units you are looking for in your answer first.
3. Then start the picket fence with the number that is given to you in the problem and is not part of a proportion. The important part is to write down the units. Picket fences are just converting units.
4. Make sure to cancel units by putting the units that are the same kiddy corner from each other in the picket fence.

Examples

1. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with NaOH how many moles of NaCl will be formed?

Step 1: Finish the chemical equation. As stated in the problem the combination of an acid and a base make a salt and water. So the products would have to be NaCl and HOH. This can also be determined because it is a double replacement reaction.

Step 2: Balance the chemical equation. Looking at each of the individual atoms you can see that everything is balanced without changing anything.

Step 3: Determine what units are required in the answer. The problem asks for the number of moles of NaCl produced.

Step 4: Start the picket fence with the number given that is not part of a proportion. This would be 25 moles of HCl.

Step 5: Cancel units and do the math. Remember that the balanced chemical equation shows the ratio of moles between all of the substances.

 25 mole HCl 1 mole NaCl = 25 moles NaCl 1 mole HCl

1. If 25 grams of magnesium combines with oxygen in a composition reaction, how many moles of magnesium oxide will be formed?

Step 1: Balance the chemical equation. Starting with the oxygen there are 2 on the reactant side and one on the product side. Thus, the product needs to be multiplied by two. This will then change the number of magnesium. So magnesium on the reactant side needs to be multiplied by two. 2 Mg + O2 ą 2 MgO

Step 2: Determine what units are required in the answer. The problem asks for the number of moles of magnesium oxide.

Step 3: Start the picket fence with the number given that is not part of a proportion. This would be 25 grams of magnesium.

Step 4: Cancel units and do the math. Remember that the balanced chemical equation shows the ratio of moles between all of the substances. Also, Remember the gram formula mass from the periodic table.

 25 gram Mg 1 mole Mg 2 mole MgO = 1.04 mole MgO 24 grams Mg 2 mole Mg

8. If 70 grams of gold III chloride decomposes into its elements, how many grams of gold will be produced?

Step 1: Finish the chemical equation. As stated in the problem this is a decomposition reaction and decomposes into its elements gold and chlorine. Remember that chlorine is a diatomic

Step 2: Balance the chemical equation. Looking at each of the individual atoms you can see that the chlorines are not balanced. The least common multiple between 2 and 3 is 6. So the AuCl3 needs to be multiplied by 2 and chlorine needs to be multiplied by 3. This changes the number of gold atoms on each side of the equation. There are 2 on the reactant side and one on the product side so the gold on the product sides needs to be multiplied by 2.

2 AuCl3 ą 2 Au + 3 Cl2

Step 3: Determine what units are required in the answer. The problem asks for the number of grams of gold produced.

Step 4: Start the picket fence with the number given that is not part of a proportion. This would be 70 grams of AuCl3.

Step 5: Cancel units and do the math. . Remember that the balanced chemical equation shows the ratio of moles between all of the substances. Also, Remember the gram formula mass from the periodic table.

 70 gram AuCl3 1 mole AuCl3 2 moles Au 197 grams Au = 45.4 grams Au 303.5 grams AuCl3 2 moles AuCl3 1 mole Au

1. Acid rain is produced when sulfur trioxide reacts with water in a composition reaction. If 15 L of sulfur trioxide is present in the air how many moles of sulfuric acid are produced?

Step 1: Finish the chemical equation. As stated in the problem this is a composition reaction, so the product will be a combination of the two reactants. In this case it will be sulfuric acid.

Step 2: Balance the chemical equation. Looking at each of the individual atoms you can see that everything is already balanced.

SO3 + H2O ą H2SO4

Step 3: Determine what units are required in the answer. The problem asks for the number of moles of acid produced.

Step 4: Start the picket fence with the number given that is not part of a proportion. This would be 15 L of SO3.

Step 5: Cancel units and do the math. . Remember that the balanced chemical equation shows the ratio of moles between all of the substances. Also, Remember the molar volume of a gas is 22 L / mole.

 15 L SO3 1 mole SO3 1 moles H2SO4 = 0.68 moles H2SO4 22 L SO3 1 moles SO3

14. By running a direct current through water, it can be decomposed into its elements. How many liters of hydrogen gas can be produced from 1500 g of water?

Step 1: Finish the chemical equation. As stated in the problem this is a decomposition reaction in which water breaks into its elements. Remember that hydrogen and oxygen are both diatomics.

Step 2: Balance the chemical equation. Looking at each of the individual atoms you can see that there is one oxygen on the reactant side and 2 on the product side so water needs to be multiplied by 2. This then changes the number of hydrogens to 4 on the reactant side and 2 on the product side. Hydrogen needs to be multiplied by 2.

2 H2O ą 2 H2 + O2

Step 3: Determine what units are required in the answer. The problem asks for the number of liters of hydrogen gas produced.

Step 4: Start the picket fence with the number given that is not part of a proportion. This would be 1500 grams of water.

Step 5: Cancel units and do the math. . Remember that the balanced chemical equation shows the ratio of moles between all of the substances. Also, Remember the molar volume of a gas is 22 L / mole.

 1500 g H2O 1 mole H2O 2 moles H2 22 L H2 =1833 L H2 18 g H2O 2 moles H2O 1 mole H2

17. How many liters of nitrogen monoxide can be made using 18 liters of oxygen in the following composition reaction?

Step 1: Finish the chemical equation. As stated in the problem this is a composition reaction, so the product will be a combination of the two reactants. In this case it will be nitrogen monoxide.

Step 2: Balance the chemical equation. Looking at each of the individual atoms you can see that there are 2 nitrogens and 2 oxygens on the reactant side and only one of each on the product side. Nitrogen monoxide needs to be multiplied by 2.

N2 + O2 ą 2 NO

Step 3: Determine what units are required in the answer. The problem asks for the number of liters of NO produced.

Step 4: Start the picket fence with the number given that is not part of a proportion. This would be 18 L of O2.

Step 5: Cancel units and do the math. . Remember that the balanced chemical equation shows the ratio of moles between all of the substances. Also, Remember the molar volume of a gas is 22 L / mole.

 18 L O2 1 mole O2 2 moles NO 22 L NO = 36 L NO 22 L O2 1 moles O2 1 mole NO