How to Balance Chemical Equations
This is part of Year 11 HSC Chemistry course under the topic of Chemical Reactions and Stoichiometry
HSC Chemistry Syllabus
- Relate stoichiometry to the law of conservation of mass in chemical reactions by investigating:
- Balancing chemical equations (ACSCH039)
- Solving problems regarding mass changes in chemical reactions (ACSCH046)
This is part of Year 11 HSC Chemistry course under the topic of Mole Concept.
Balancing Chemical Equations
Chemical equations show how reactants combine to form products in a chemical reaction. For a chemical equation to be valid, it must also be balanced. This means that the number of atoms of each element must be the same on both sides of the equation
Balancing equations reflect the law of conservation of mass: atoms are neither created nor destroyed during a chemical reaction.
Reactants and Products
- Reactants are written on the left-hand side of the arrow
-
Products are written on the right-hand side of the arrow
- The arrow means "reacts to form"
Coefficients and Subscripts
- Subscripts show how many atoms are in a molecule and are fixed numbers
- Coefficients are numbers written in front of chemical formulas and multiply the entire formula
- Only coefficients are changed when balancing equations.
Practice Question 1 (Explanations in video)
Formation of water
Hydrogen gas reacts with oxygen gas to form water
Explanation:
$$H_2(g) + O_2(g) \rightarrow H_2O$$
Atom count:
- Reactants (left side of the arrow): 2 hydrogen atoms, 2 oxygen atoms
- Products (right side of the arrow): 2 hydrogen atoms, 1 oxygen atom
Because the number of oxygen atoms is not equal on the two sides of the equation is described to be unbalanced.
To balance the number of oxygen atoms, place a coefficient of 2 in front of water:
$$H_2(g) + O_2(g) \rightarrow \textbf{2}H_2O$$
Now counting atoms again:
- Oxygen: 2 on each side
- Hydrogen: 4 on the product side, 2 on the reactant side
To balance hydrogen, place a coefficient of 2 in front of hydrogen gas:
$$\textbf{2}H_2(g) + O_2(g) \rightarrow \textbf{2}H_2O(l)$$
Alternatively
If hydrogen is balanced first, then a fraction of a half will be required in front of the oxygen.
$$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O$$
While in the equation above, the atoms are balanced on both sides of the equation, fractional coefficients are not usually left in final answers. Multiply the entire equation by 2 to remove the fractions.
$$\textbf{2}H_2(g) + O_2(g) \rightarrow \textbf{2}H_2O(l)$$
Practice Question 2 (Explanations in video)
Formation of Ammonia
Nitrogen gas reacts with hydrogen gas to form ammonia.
Explanation:
$$N_2(g) + H_2(g) \rightarrow NH_3(g)$$
Nitrogen is easier to balance first. Add a coefficient of 2 to ammonia:
$$N_2(g) + H_2(g) \rightarrow \textbf{2}NH_3(g)$$
Counting the atoms:
-
Products: 6 hydrogen atoms
-
Reactants: 2 hydrogen atoms per H₂ molecule
Add a coefficient of 3 to hydrogen gas to balance the hydrogen atoms:
$$N_2(g) + \textbf{3}H_2(g) \rightarrow \textbf{2}NH_3(g)$$
The equation is now balanced.
Practice Question 3 (Explanations in video)
Ion Exchange Reaction
A polyatomic ion is a group of atoms that stay together during a reaction. When possible, polyatomic ions should be balanced as a single unit
Explanation:
$$K_2SO_4(aq) + CaBr_2(aq) \rightarrow CaSO_4(aq) + KBr$$
The sulfate ion (`SO_4^{2–}`) appears unchanged on both sides, it can be balanced first since it is already in ratio.
Next balance potassium by placing a 2 in front of to the potassium bromide
$$K_2SO_4(aq) + CaBr_2(aq) \rightarrow CaSO_4(aq) + \textbf{2}KBr$$
All atoms are now balanced
Practice Question 4 (Explanations in video)
Acid Base Reaction
Explanation:
$$Pb(OH)_2(aq) + HCl(aq) \rightarrow PbCl_2(aq) + H_2O(l)$$
There are two oxygen atoms in the lead hydroxide but only one in water, so place a 2 in front of the water
$$Pb(OH)_2(aq) + HCl(aq) \rightarrow PbCl_2(aq) + \textbf{2}H_2O(l)$$
Now balance the remaining hydrogen and chlorine by adding a 2 in front of the hydrochloric acid:
$$Pb(OH)_2(aq) + \textbf{2}HCl(aq) \rightarrow PbCl_2(aq) + \textbf{2}H_2O(l)$$
Practice Question 5 (Explanations in video)
Incomplete Combustion Reaction
Incomplete combustion reactions can have more than one correct balanced equation. This is because carbon atoms from the fuel may form both carbon dioxide (`CO_2`) and carbon monoxide (`CO`).
Explanation:
$$C_8H_18(l) + O_2(g) \rightarrow CO_2(g) + CO(g) + H_2O(l)$$
Hydrogen must be balanced first because it appears in only one reactant and one product. Balancing hydrogen requires 9 water molecules
$$C_8H_18(l) + O_2(g) \rightarrow CO_2(g) + CO(g) + \textbf{9}H_2O(l)$$
The 8 remaining carbons can be split between `CO_2` and `CO` in any ratio that equals 8.
e.g.
$$C_8H_{18}(l) + O_2(g) \rightarrow \textbf{4}CO_2(g) + \textbf{4}CO(g) + \textbf{9}H_2O(l)$$
$$C_8H_{18}(l) + O_2(g) \rightarrow \textbf{3}CO_2(g) + \textbf{5}CO(g) + \textbf{9}H_2O(l)$$
$$C_8H_{18}(l) + O_2(g) \rightarrow \textbf{2}CO_2(g) + \textbf{6}CO(g) + \textbf{9}H_2O(l)$$
etc.
Once carbon is balanced, oxygen is balanced last
$$C_8H_{18}(l) + \frac{3}{2}O_2(g) \rightarrow \textbf{4}CO_2(g) + \textbf{4}CO(g) + \textbf{9}H_2O(l)$$
$$C_8H_{18}(l) + O_2(g) \rightarrow \textbf{3}CO_2(g) + \textbf{5}CO(g) + \textbf{9}H_2O(l)$$
$$C_8H_{18}(l) + \frac{1}{2}O_2(g) \rightarrow \textbf{2}CO_2(g) + \textbf{6}CO(g) + \textbf{9}H_2O(l)$$
Observe that different combinations of `CO_2` and `CO` lead to different, but still valid, balanced equations
Practice Question 6 (Explanations in video)
Acid-carbonate reaction
Explanation:
$$HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$$
Acid-carbonate reactions always produce:
- A salt
- Water
- Carbon Dioxide
Balance hydrogen and chlorine first by placing a 2 in front of HCl
$$\textbf{2}HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$$