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Balancing equations

Balancing equations

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Explainer Video

Tutor: Aimee LeBrun

Summary

Balancing equations

In a nutshell

Chemical equations are balanced to ensure there is the same number of each type of atom on either side of the reaction. 



Balancing an equation 

Numbers are placed in front of reactants and products to balance equations and ensure that there are equal numbers of each type of atom on either side of the equation. 


procedure

1.

Count the number of atoms on each side of the equation. If the numbers are unequal the equation needs to be balanced.

2.

To balance the equation, add numbers in front of the substances to make the number of each type of atom equal on either side of the reaction.

3.

Before the equation is finished and balanced, be sure to count the numbers of each type of atoms on either side again to be absolutely sure. 



Example 

Balance the following equation. 


Al + H2SO4 → Al2(SO4)3 + H2Al \, + \, H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_2Al+H2​SO4​→Al2​(SO4​)3​+H2​


First, count the number of each type of atom on either side of the equation:  


​​type of atom

left-hand side

right-hand side

AlAlAl​​

​111​​

​​2​​2​​2​​​

HHH​​

​​​2​​2​​2​​

​2​​2​​2​​​​

SSS​​

​1​​1​​1​​​​

​3​​3​​3​​​​

OOO​​

444​​

121212​​

​

Add a two in front of AlAlAl on the left-hand side to balance the number of aluminium atoms:


2Al + H2SO4 → Al2(SO4)3 + H22Al \, + \, H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_22Al+H2​SO4​→Al2​(SO4​)3​+H2​​​


Recount the number of atoms on either side of the reaction:


2Al + H2SO4 → Al2(SO4)3 + H22Al \, + \, H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_22Al+H2​SO4​→Al2​(SO4​)3​+H2​​​


​​type of atom

left-hand side

right-hand side

AlAlAl​​

222​​

222​​

HHH​​

222​​

222​​

SSS​​

111​​

333​​

OOO​​

444​​

121212​​


Add a three in front of the H2SO4H_2SO_4H2​SO4​ on the left-hand side to balance the number of oxygen and sulfur atoms :


2Al + 3H2SO4 → Al2(SO4)3 + H22Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_22Al+3H2​SO4​→Al2​(SO4​)3​+H2​​​


Recount the number of atoms on either side of the reaction:


2Al + 3H2SO4 → Al2(SO4)3 + H22Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_22Al+3H2​SO4​→Al2​(SO4​)3​+H2​​​


​​type of atom

left-hand side

right-hand side

AlAlAl​​

222​​

222​​

HHH​​

666​​

222​​

SSS​​

333​​

333​​

OOO​​

121212​​

121212​​


The number of hydrogen atoms on the left-hand side has increased, add a three in front of the H2H_2H2​​ on the right-hand side to balance the equation:


2Al + 3H2SO4 → Al2(SO4)3 + 3H22Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, 3H_22Al+3H2​SO4​→Al2​(SO4​)3​+3H2​​​


Finally, be sure to do a final count of each type of atoms either side of the equation to ensure that your equation is balanced correctly. 


2Al + 3H2SO4 → Al2(SO4)3 + 3H22Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, 3H_22Al+3H2​SO4​→Al2​(SO4​)3​+3H2​​​


​​type of atom

left-hand side

right-hand side

AlAlAl​​

222​​

222​​

HHH​​

666​​

666​​

SSS​​

333​​

333​​

OOO​​

121212​​

121212​​

​

The numbers of each type of atom on either side of the equation are equal and the equation is now balanced; 2Al + 3H2SO4 → Al2(SO4)3 + 3H2‾\underline{2Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, 3H_2}2Al+3H2​SO4​→Al2​(SO4​)3​+3H2​​.​



Finding masses from balanced equations 

The stoichiometry discovered when balancing an equation can be used alongside the reacting masses to find the masses of products. 


procedure 

​​1.

Use the mass of a reactant to find the number of moles of the substance:


​n = mMrn \, = \, \frac{m}{M_r}n=Mr​m​​​

2.

The number of moles of the product can be found using the calculated number of moles of the reactant in step 1 along with the mole ratio from the balanced equation. 

3.

Use the number of moles of the product to find the amount of mass that should be produced using:


​m = n × Mrm \, = \, n \, \times \, M_rm=n×Mr​​​


 

Ionic equations  

Given the full balanced equation of a reaction, ionic equations can be constructed. 


PROCEDURE 

1.
Distinguish between molecules and ionic compounds.
2.
Distinguish between ionic compounds with ions that can move and substances that are not ions in solution.
3.
Re-write the symbol equation, separating the ions but keeping any solids or substances that are not ions in solution the same.
4.
Cross out ions that are the same on both sides. These are called spectator ions, they are not involved in the reaction.
5.
The ionic equation is left. 


Example

Write the ionic equation for the reaction below.


HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)HCl(aq) \, + \, NaOH(aq) \, \rightarrow \, NaCl(aq) \, + \, H_2O(l)HCl(aq)+NaOH(aq)→NaCl(aq)+H2​O(l)


First, identify any substance that will not break into ions in solution:


H2O(l)H_2O(l)H2​O(l)

​​

Next, write out the ions present in this reaction:


H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → Na+(aq) + Cl− (aq)+ H2O(l)H^+(aq) \, + \, Cl^-(aq) \, + \, Na^+(aq) \, + \, OH^-(aq) \, \rightarrow \, Na^+(aq) \, + \, Cl^- \,(aq) + \, H_2O(l)H+(aq)+Cl−(aq)+Na+(aq)+OH−(aq)→Na+(aq)+Cl−(aq)+H2​O(l)​​


Cross out the ions that are present on both sides of the equation:


H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → Na+(aq) + Cl−(aq) + H2O(l)H^+(aq) \, + \, \cancel{Cl^-(aq)} \, + \, \cancel{Na^+(aq)} \, + \, OH^-(aq) \, \rightarrow \, \cancel{Na^+(aq)} \, + \, \cancel{Cl^-(aq)} \, + \, H_2O(l)H+(aq)+Cl−(aq)​+Na+(aq)​+OH−(aq)→Na+(aq)​+Cl−(aq)​+H2​O(l)​​


Write the ions that are left:


H+(aq) + OH−(aq) → H2O(l)H^+(aq) \, + \, OH^-(aq) \, \rightarrow \, H_2O(l)H+(aq)+OH−(aq)→H2​O(l)​​


The ionic equation for this reaction is H+(aq) + OH−(aq) → H2O(l)‾\underline{H^+(aq) \, + \, OH^-(aq) \, \rightarrow \, H_2O(l)}H+(aq)+OH−(aq)→H2​O(l)​. 



State symbols 

State symbols are used to tell you what state of matter a substance is. State symbols can be used to explain what is happening during a reaction. 


STATE

SYMBOL

​ SolidSolidSolid​​
​ (s)(s)(s)​​
​ LiquidLiquidLiquid​​
​ (l)(l)(l)​​
​ GasGasGas​​
​ (g)(g)(g)​​
​ Aqueous solutionAqueous ~solutionAqueous solution​​
​(aq)(aq)(aq)


State symbols signify what type of products are produced during reactions. 


Precipitate reactions 

A precipitate reaction occurs when two aqueous substances react and one of the products is a solid. 


Neutralisation reactions 

Acids and bases react together to form salt products along with carbon dioxide and water. 


Displacement reactions 

More reactive metals can displace less reactive metals in a displacement reaction. 


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FAQs - Frequently Asked Questions

What do state symbols describe?

State symbols signify what state of matter a substance is.

What are spectator ions?

Spectator ions are ions that are not involved in a reaction.

How are chemical equations balanced?

Numbers are placed in front of reactants and products to balance equations and ensure that there are equal numbers of each type of atom on either side of the equation.

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