Balanced chemical equations provide a lot of useful information about chemical reactions – to describe a reaction properly you need to be capable of correctly balancing equations. Work through the examples below to revise this important topic, then try the practice questions at the bottom!
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The reaction between solid calcium carbonate, CaCO3, and a solution of hydrochloric acid, HCl, produces a solution of calcium chloride, CaCl2, water, H2O, and carbon dioxide gas, CO2. This reaction can be represented by the following equation:
This is known as a full equation because it shows the full chemical formula of all of the reactants (on the left–hand side) and all of the products (on the right–hand side).
The arrow in the equation represents the transformation that occurs during the chemical reaction.
The numbers in front of the formulae of the reactants and products in a balanced equation represent the stoichiometry of the reaction. This is the ratio in which reactants react with each other and products are produced. In the example above, one mole of calcium carbonate would react completely with two moles of hydrochloric acid to produce one mole of calcium chloride, one mole of water and one mole of carbon dioxide. It is the stoichiometry of the reaction that you adjust when balancing equations.
Finally, state symbols indicate the physical state of reactants and products during the chemical reaction: (s) represents a solid, (l) represents a liquid, (g) represents a gas and (aq) represents an aqueous solution (a compound dissolved in water).
Balancing Equations
Atoms are not lost or gained in chemical reactions; they are simply rearranged to form different substances. Therefore, the total number of each type of atom represented on the left–hand side of an equation must equal the total number of each type of atom represented on the right–hand side of an equation. In the full equation above, the number of atoms of each element is the same on both sides of the equation, so the equation is balanced.
| Element | Number of Atoms in Reactants | Number of Atoms in Products |
|---|---|---|
| calcium, Ca | 1 | 1 |
| Carbon, C | 1 | 1 |
| oxygen, O | 3 | 3 |
| hydrogen, H | 2 | 2 |
| chlorine, Cl | 2 | 2 |
If the equation is not balanced, you must change the stoichiometric ratios in the equation until it is balanced.
Example Question 1
Write a balanced full equation for the reaction between sodium metal and oxygen to produce sodium oxide.
Correctly balancing equations requires you to use your knowledge of the periodic table and chemical formulae, structure and bonding to deduce the chemical formulae of the reactants and products. You can assume that all substances are in their standard states unless otherwise stated.
Formulae of Reactants
sodium Na(s)
oxygen O2(g)
Formula of Product
sodium oxide Na2O(s)
(Sodium oxide is an ionic compound containing Na+ and O2– ions.)
Na(s) + O2(g) → Na2O(s)
Next, check the number of each type of atom on each side of the equation.
| Element | Number of Atoms in Reactants | Number of Atoms in Products |
|---|---|---|
| sodium, Na | 1 | 2 |
| oxygen, O | 2 | 1 |
The equation is not balanced. Both the number of oxygen atoms on the right-hand side and the number of sodium atoms on the left hand side are wrong. Start by correcting the number of oxygen atoms on the right – increasing the number of molecules of Na2O will also change the amount of sodium required on the left-hand side, which can be adjusted later.
Na(s) + O2(g) → 2Na2O(s)
| Element | Number of Atoms in Reactants | Number of Atoms in Products |
|---|---|---|
| sodium, Na | 1 | 4 |
| oxygen, O | 2 | 2 |
Four atoms of sodium are now needed on the left–hand side of the equation.
4Na(s) + O2(g) → 2Na2O(s)
| Element | Number of Atoms in Reactants | Number of Atoms in Products |
|---|---|---|
| sodium, Na | 4 | 4 |
| oxygen, O | 2 | 2 |
The equation is now balanced: 4Na(s) + O2(g) → 2Na2O(s)
Ionic Equations
In some reactions, there are ions present that do not take part in the overall reaction. These are known as spectator ions (ions that have the same formula, charge and physical state on both sides of the equation).
An ionic equation can be used to show only the ions that take part in the reaction.
The steps to writing an ionic equation are outlined below.
- Start with the full balanced equation for the reaction.
CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) - Identify any ions that are present in each reactant and product.
| Reactant/Product | Ions present |
|---|---|
| CaCO3(s) | Ca2+(s) and CO32–(s) |
| HCl(aq) | H+(aq) and Cl–(aq) |
| CaCl2(aq) | Ca2+(aq) and Cl–(aq) |
| H2O(l) | none (covalent compound) |
| CO2(g) | none (covalent compound) |
- Write the equation out showing all ions that are present.
Ca2+(s) + CO32–(s) + 2H+(aq) + 2Cl–(aq) → Ca2+(aq) + 2Cl–(aq)+ H2O(l) + CO2(g) - Identify the spectator ions.
These are ions that have not taken part in the overall reaction and remain unchanged. The spectator ions can be cancelled out of the equation.
Ca2+(s) + CO32–(s) + 2H+(aq) +2Cl–(aq)→ Ca2+(aq) +2Cl–(aq)+ H2O(l) + CO2(g) - Rewrite the equation without the spectator ions.
At this stage you can write the full chemical formula for compounds where all ions are still present in the equation.
CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
Example Question 2
Write a balanced ionic equation for the reaction between potassium iodide solution and silver nitrate solution to form a precipitate of silver iodide and soluble potassium nitrate.
Start by writing a balanced, full equation:
Formulae of Reactants
potassium iodide KI(aq)
silver nitrate AgNO3(aq)
Formulae of Products
silver iodide AgI(s)
potassium nitrate KNO3(aq)
Balanced full equation:
KI(aq) + AgNO3(aq) → AgI(s) + KNO3(aq)
In this case, it is easy to see by inspection that the equation is balanced. Next, identify the ions present:
| REACTANT/PRODUCT | IONS PRESENT |
|---|---|
| KI (aq) | K+ (aq) and I– (aq) |
| AgNO3 (aq) | Ag+(aq) and NO3–(aq) |
| AgI (s) | Ag+ (s) and I– (s) |
| KNO3 (aq) | K+ (aq) and NO3– (aq) |
Next, write an equation with ions:
K+(aq) + I–(aq) + Ag+(aq) + NO3–(aq) → Ag+(s) + I–(s) + K+(aq) + NO3–(aq)
Then, identify the spectator ions and cancel them out:
K+(aq) + I–(aq) + Ag+(aq) + NO3–(aq) → Ag+(s) + I–(s) + K+(aq) + NO3–(aq)
This leaves a balanced ionic equation of:
I–(aq) + Ag+(aq) → AgI(s)
Practice Questions
1. Write a balanced full equation for the reaction between magnesium metal and oxygen to produce magnesium oxide.
2. Write a balanced full equation for the reaction between lithium hydroxide solution and sulfuric acid to produce soluble lithium sulfate and water.
3. Write a balanced full equation for the reaction between solid aluminium oxide and phosphoric acid to produce insoluble aluminium phosphate and water.
4. Write a balanced ionic equation for the reaction between hydrochloric acid and sodium hydroxide solution to produce sodium chloride and water.
5. Write a balanced ionic equation for the reaction between magnesium metal and iron(II) nitrate solution to produce magnesium nitrate solution and iron metal.
6. Write a balanced ionic equation for the reaction between barium chloride solution and sodium sulfate solution to produce a precipitate of barium sulfate and aqueous sodium chloride.
Answers