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.

$Al \, + \, H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_2$

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

type of atom	left-hand side	right-hand side
$Al$	$1$	$2$
$H$	$2$	$2$
$S$	$1$	$3$
$O$	$4$	$12$

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

$2Al \, + \, H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_2$ 

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

$2Al \, + \, H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_2$ 

type of atom	left-hand side	right-hand side
$Al$	$2$	$2$
$H$	$2$	$2$
$S$	$1$	$3$
$O$	$4$	$12$

Add a three in front of the $H_2SO_4$ on the left-hand side to balance the number of oxygen and sulfur atoms :

$2Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_2$

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

$2Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, H_2$ 

type of atom	left-hand side	right-hand side
$Al$	$2$	$2$
$H$	$6$	$2$
$S$	$3$	$3$
$O$	$12$	$12$

The number of hydrogen atoms on the left-hand side has increased, add a three in front of the $H_2$  on the right-hand side to balance the equation:

$2Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, 3H_2$

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.

$2Al \, + \, 3H_2SO_4 \, \rightarrow \, Al_2(SO_4)_3 \, + \, 3H_2$ 

type of atom	left-hand side	right-hand side
$Al$	$2$	$2$
$H$	$6$	$6$
$S$	$3$	$3$
$O$	$12$	$12$

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

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

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

$n \, = \, \frac{m}{M_r}$

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.

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

$m \, = \, n \, \times \, M_r$

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.

$HCl(aq) \, + \, NaOH(aq) \, \rightarrow \, NaCl(aq) \, + \, H_2O(l)$

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

$H_2O(l)$

Next, write out the ions present in this reaction:

$H^+(aq) \, + \, Cl^-(aq) \, + \, Na^+(aq) \, + \, OH^-(aq) \, \rightarrow \, Na^+(aq) \, + \, Cl^- \,(aq) + \, H_2O(l)$

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

$H^+(aq) \, + \, \cancel{Cl^-(aq)} \, + \, \cancel{Na^+(aq)} \, + \, OH^-(aq) \, \rightarrow \, \cancel{Na^+(aq)} \, + \, \cancel{Cl^-(aq)} \, + \, H_2O(l)$

Write the ions that are left:

$H^+(aq) \, + \, OH^-(aq) \, \rightarrow \, H_2O(l)$

The ionic equation for this reaction is $\underline{H^+(aq) \, + \, OH^-(aq) \, \rightarrow \, H_2O(l)}$ .

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
$Solid$	$(s)$
$Liquid$	$(l)$
$Gas$	$(g)$
$Aqueous ~solution$	$(aq)$

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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