Balanced Equations and Related Calculations

This lesson covers: 

  1. How to balance chemical equations
  2. State symbols
  3. How to write ionic equations
  4. Calculating amounts of reactants and products 
  5. Limiting reactants
  6. Calculating amounts of products based on the limiting reactant

Balancing chemical equations

A balanced chemical equation has equal numbers of atoms for each element on both sides of the equation.

To balance an equation:

  1. Write out the unbalanced equation.
  2. Count the number of atoms of each element on both sides.
  3. Add coefficients before formulas to balance atom numbers.
  4. Check if atom numbers are equal on both sides.


You cannot alter the chemical formulas themselves.

Balanced equations obey the law of conservation of mass, showing that the same mass is retained before and after a reaction.

Worked example 1 - Balancing the complete combustion of propane

Balance the equation for the complete combustion of propane (C3H8) with oxygen to form carbon dioxide and water.

Step 1: Write out the unbalanced equation

C3H8 + O2 ➔ CO2 + H2O


Step 2: Count the number of atoms of each element on both sides

C3H8O2➔CO2H2O
C31
H82
O221


Step 3: Add coefficients before formulas to balance atom numbers

  • Start with carbon: To balance 3 C atoms in propane, add a coefficient of 3 to CO2.
  • Next, balance hydrogen: To balance 8 H atoms, H2O needs a coefficient of 4 (since each molecule has 2 H atoms).
  • Finally, balance oxygen: There are now 3(2) + 4 = 10 O atoms required for CO2 and H2O. Since O2 has 2 O atoms, you need 5 molecules of O2 to provide 10 O atoms.


Step 4: Check if atom numbers are equal on both sides

The balanced equation is:

C3H8 + 5O2 ➔ 3CO2 + 4H2O

C3H85O2➔3CO24H2O
C33
H88
O1064

This balances the equation, ensuring the law of conservation of mass is satisfied.

State symbols

State symbols are written after chemical formulas or names of substances in chemical equations. They indicate the physical state of each reactant and product.

Common state symbols are:

  • (s) - solid
  • (l) - liquid
  • (g) - gas
  • (aq) - aqueous (dissolved in water)


For example:

CaCO3(s) + 2HCl(aq) ➔ CaCl2(aq) + H2O(l) + CO2(g)


The state symbols show that:

  • Calcium carbonate (CaCO3) is a solid.
  • Hydrochloric acid (HCl) and calcium chloride (CaCl2) are aqueous solutions.
  • Water (H2O) is a liquid
  • Carbon dioxide (CO2) is a gas.

Ionic equations for reactions in aqueous solution

Many reactions occur between ions dissolved in aqueous solution. For these reactions, you can write an ionic equation which shows only the particles directly involved in the reaction.

To write an ionic equation:

  1. Write out the full balanced equation.
  2. Split soluble ionic compounds into their component ions.
  3. Cancel out any ions that appear unchanged on both sides (called spectator ions).
  4. Check ionic charges balance on both sides.

Worked example 2 - Writing an ionic equation for the precipitation of silver chloride

A solution of silver nitrate reacts with sodium chloride to form silver chloride precipitate and sodium nitrate solution.

Write the ionic equation.


Step 1: Write out the full balanced equation including state symbols

AgNO3(aq) + NaCl(aq) ➔ AgCl(s) + NaNO3(aq)


Step 2: Split soluble ionic compounds into their component ions

Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) ➔ AgCl(s) + Na+(aq) + NO3−(aq)


Step 3: Cancel out spectator ions

Ag+(aq) + Cl−(aq) ➔ AgCl(s)


Step 4: Check ionic charges balance on both sides

Charges balance: +1 from Ag+ and -1 from Cl− equals 0, indicating a neutral compound is formed.

Using balanced equations to calculate masses

Balanced equations show molar ratios between reactants and products. By rearranging ratios and using molar masses, you can calculate:

  • Mass of product from given mass of reactant.
  • Mass of reactant needed to produce a given mass of product.


The steps to calculate reacting masses are:

  1. Write out the balanced equation.
  2. Convert given mass into moles.
  3. Use the mole ratio from the equation to calculate moles of the desired species.
  4. Convert moles of the desired species into mass.

Worked example 3 - Calculating mass of product

50.0 g of calcium carbonate reacts with excess hydrochloric acid to produce calcium chloride, water, and carbon dioxide.

Calculate the mass (in g) of carbon dioxide produced.


Step 1: Write out the balanced equation

CaCO3(s) + 2HCl(aq) ➔ CaCl2(aq) + H2O(l) + CO2(g)


Step 2: Calculate moles of CaCO3

moles of CaCO3​=Mr​mass​=100.150.0​=0.4995 mol


Step 3: Calculate moles of CO2 produced

CaCO3 : CO2 mole ratio = 1:1

Moles of CO2 = 0.4995 mol


Step 4: Calculate mass of CO2

mass of CO2 = moles x Mr = 0.4995 x 44.0 = 22.0 g

Worked example 4 - Calculating mass of reactant

Calculate the mass (in g) of aluminium (Al) required to react completely with oxygen (O2) to produce 100.0 g of aluminium oxide (Al2O3).


Step 1: Write out the balanced equation

4Al(s) + 3O2(g) ➔ 2Al2O3(s)


Step 2: Calculate moles of Al2O3

moles of Al2​O3​=Mr​ mass​=102.0100.0​=0.980 mol


Step 3: Calculate moles of Al required

Al2O3 : Al mole ratio = 2:4 = 1:2

Moles of Al required = 0.980 × 2 = 1.960 mol


Step 4: Calculate mass of Al

mass of Al = moles x Ar = 1.960 x 27.0 = 52.9 g

The limiting reactant limits the amount of product

When two or more reactants are mixed together, one reactant will always be completely used up before the others.

  • The reactant that runs out first is called the limiting reactant.
  • Any reactants present in excess are called excess reactants.
  • The amount of limiting reactant present sets an upper limit on how much product can be formed.


For example, in the reaction:

A + B ➔ C


If 1 mole of A reacts with 2 moles of B, then A will be the limiting reactant as B is present in excess.

So if the reaction proceeds to completion, 1 mole of A will produce 1 mole of C, leaving 1 mole of B unreacted.

How to identify the limiting reactant

There are two steps to determining which reactant is limiting:

  1. Calculate the number of moles of each reactant present.
  2. Compare the number of moles of each reactant to the stoichiometric ratios in the balanced chemical equation:
  • The reactant for which the available number of moles is lower relative to the stoichiometric ratio is the limiting reactant.
  • The other reactants, for which there are excess moles beyond the stoichiometric ratio, are in excess.


For example, consider the reaction between hydrogen and oxygen to form water:

2H2(g) + O2(g) ➔ 2H2O(l)


According to the balanced equation, the stoichiometric ratio is 2 moles of H2 per 1 mole of O2.


If equal 1 mole quantities of H2 and O2 react, there would only be 1 mole of H2 per 1 mole O2, instead of the required 2:1 ratio. Since the amount of H2 is lower than the stoichiometric ratio relative to O2, hydrogen would be the limiting reactant.

Worked example 5 - Determining the limiting reactant

23.0 g of C2H5OH undergoes complete combustion with 50.0 g of O2. The balanced equation is: 

C2H5OH(l) + 3O2(g) ➔ 2CO2(g) + 3H2O(g)

Determine the limiting reactant.


Step 1: Calculate number of moles of each reactant

Moles of C2​H5​OH =Mr ​m​=46.023.0​=0.50 mol

Moles of O2​ =Mr ​m​=32.050.0​=1.56 mol


Step 2: Compare to stoichiometric ratio (1:3)

For 0.50 mol of C2H5OH, 0.50 x 3 = 1.50 mol of O2 is required


Step 3: Compare number of O2 moles required vs. present

Moles of O2 required = 1.50 mol

Moles of O2 present = 1.56 mol


Step 4: Determine the limiting reactant

1.56 > 1.50 so O2 is in excess and C2H5OH is the limiting reactant.

Basing calculations on limiting reagent

The maximum amounts of products formed depends on the limiting reactant:

  • Amounts of products can only be calculated based on moles of limiting reactant.
  • Excess reactants does not affect product amounts


For example, in the reaction:

C2H5OH(l)+3O2(g) ➔ 2CO2(g)+3H2O(g)


0.50 mol C2H5OH can produce:

  • 0.50 × 2 = 1.0 mol of CO2
  • 0.50 × 3 = 1.50 mol H2O


So the maximum amounts of carbon dioxide and water produced are 1.0 mol and 1.50 mol respectively, regardless of O2 in excess.

Performing calculations based on limiting reagent is essential to determine actual yields.

Worked example 6 - Using the limiting reactant to calculate mass of product

Calculate the mass (in g) of sodium chloride (NaCl) that can be produced when 200 cm3 of 1.0 mol dm-3 hydrochloric acid (HCl) reacts with 150 cm3 of 1.2 mol dm-3 sodium hydroxide (NaOH). The balanced equation is:

HCl(aq) + NaOH(aq) ➔ NaCl(aq) + H2O(l)


Step 1: Conversion of cm3 into dm3

To convert from cm3 into dm3, divide by 1,000

200 cm3 = 0.200 dm3

150 cm3 = 0.150 dm3


Step 2: Calculate number of moles of each reactant

Moles of HCl = c × V = 1.0 x 0.200 = 0.200 mol

Moles of NaOH = c x V = 1.2 x 0.150 = 0.180 mol


Step 3: Compare to stoichiometric ratio (1:1)

For 0.200 mol of HCl, 0.200 mol of NaOH is required


Step 4: Compare number of NaOH moles required vs. present

Moles of NaOH required = 0.200 mol

Moles of NaOH present = 0.180 mol


Step 5: Determine the limiting reactant

0.180 < 0.200 so NaOH is the limiting reactant and HCl is in excess.


Step 6: Calculate number of moles of NaCl formed

NaCl : NaOH mole ratio = 1:1

Moles of NaCl = 0.180 mol


Step 7: Calculate mass of NaCl formed

m = n x Mr = 0.180 x 58.5 = 10.5 g

Therefore, 10.5 g of sodium chloride is produced with sodium hydroxide acting as the limiting reactant.

Worked example 7 - Using the limiting reactant to calculate mass of product

Calculate the mass (in g) of titanium metal (Ti) that can be produced when 500 g of titanium tetrachloride (TiCl4) reacts with 100 g of magnesium metal (Mg). The balanced equation is:

TiCl4(l) + 2Mg(s) ➔ Ti(s) + 2MgCl2(s)


Step 1: Calculate number of moles of each reactant

Moles of TiCl4​ =Mr ​m​=189.9500​=2.63 mol

Moles of Mg =Ar ​m​=24.3100​=4.12 mol


Step 2: Compare to stoichiometric ratio (1:2)

For 2.63 mol of TiCl4, 2.63 x 2 = 5.26 mol of Mg is required


Step 3: Compare number of Mg moles required vs. present

Moles of Mg required = 5.26 mol

Moles of Mg present = 4.12 mol


Step 4: Determine the limiting reactant

4.12 < 5.26 so Mg is the limiting reactant and TiCl4 is in excess.


Step 5: Calculate number of moles of Ti formed

Ti : Mg mole ratio = 1:2

Moles of Ti =24.12​=2.06 mol


Step 6: Calculate mass of Ti formed

m = n x Ar = 2.06 x 47.9 = 98.6 g

Therefore, 98.6 g of titanium metal can be produced, with magnesium acting as the limiting reactant.