View source for Balanced Symbol Equation

==Key Stage 3==
===Meaning===
A [[Balanced Symbol Equation]] is a way to show the [[reactant]]s and [[product]]s in a [[Chemical Reaction|chemical reaction]] using the [[Chemical Formula|chemical formulae]] of the [[reactant]]s and [[product]]s.

===About Balanced Symbol Equations===
: [[Balanced Symbol Equation]]s shows the number of [[molecule]]s of each [[chemical]] and the numbers of each [[atom]] within those [[chemical]]s.
: [[Balanced Symbol Equation]]s can be used to find the quantity of each [[chemical]] needed to have a complete [[Chemical Reaction|chemical reaction]].

{| class="wikitable"
|+ The [[Word Equation]] has been given above each Balanced Symbol Equation.
|-
|[[File:2H2+O2.png|center|500px]]
|-
| style="height:20px; width:500px; text-align:center;" |This shows there are two H<sub>2</sub> [[molecule]]s and one O<sub>2</sub> [[molecule]] needed to make two [[molecule]]s of H<sub>2</sub>O.
|-
|[[File:CH4+2O2.png|center|500px]]
|-
| style="height:20px; width:500px; text-align:center;" |This shows there is one CH<sub>4</sub> [[molecule]] and two O<sub>2</sub> [[molecule]]s needed to make one [[molecule]] of CO<sub>2</sub> and two [[molecule]]s of H<sub>2</sub>O.
|-
|[[File:2Mg+O2.png|center|500px]]
|-
| style="height:20px; width:500px; text-align:center;" |This shows there are two [[atom]]s of Mg and one  [[molecule]] of O<sub>2</sub> needed to make two [[molecule]]s of MgO.
|-
|[[File:2HCl+Li2O.png|center|500px]]
|-
| style="height:20px; width:500px; text-align:center;" |This shows there are two HCl [[molecule]]s and one Li<sub>2</sub>O [[molecule]] needed to make two [[molecule]]s of LiCl and one [[molecule]] of H<sub>2</sub>O.

|}

===Balancing Equations===
: You may know the [[Chemical Formula|chemical formulae]] of the [[reactant]]s and [[product]]s but not know the number of each [[chemical]] needed. For that you must '''balance the equation'''.
: To '''balance and equation''' you need the same number of each type of [[atom]] on the left as you do on the right.

{| class="wikitable"
|+Lithium + Fluorine → Lithium Fluoride
|-
| style="height:20px; width:200px; text-align:center;" |Step 1
| style="height:20px; width:200px; text-align:center;" |Step 2
| style="height:20px; width:200px; text-align:center;" |Step 3
|-
|[[File:Li+FStep1.png|center|200px]]
|[[File:Li+FStep2.png|center|200px]]
|[[File:Li+FStep3.png|center|200px]]
|-
| style="height:20px; width:200px; text-align:left;" |
: _Li + _F<sub>2</sub> → _LiF
: Reactants = Products
: 1Li = 1Li
: 2F ≠ 1F
We need 2 [[Fluorine]] [[atom]]s in the [[product]]s.
| style="height:20px; width:200px; text-align:left;" |
: _Li + _F<sub>2</sub> → 2LiF
: Reactants = Products
: 1Li ≠ 2Li
: 2F = 2F
There are now too many [[Lithium]] [[atom]]s in the [[product]]s so we need 2 in the [[reactant]]s.
| style="height:20px; width:200px; text-align:left;" |
: 2Li + _F<sub>2</sub> → 2LiF
: Reactants = Products
: 2Li = 2Li
: 2F = 2F
This equation is '''balanced''':

: 2Li + F<sub>2</sub> → 2LiF
|}

'''Aluminium + Chlorine → Aluminium Chloride'''
: _Al + _Cl<sub>2</sub> → _AlCl<sub>3</sub>
: Reactants = Products
: 1Al = 1Al
: 2Cl ≠ 3Cl
We need the same number of [[Chlorine]] [[atom]]s on both sides. To do that we can find the Lowest Common Multiple. In this case it is 6, so we times the left 2 [[Chlorine]]s by 3 and the right 3 [[Chlorine]]s by 2.

: _Al + 3Cl<sub>2</sub> → 2AlCl<sub>3</sub>
: Reactants = Products
: 1Al ≠ 2Al
: 6Cl = 6Cl
There are now too many [[Aluminium]] [[atom]]s on the right, so we need to multiply the left [[Aluminium]] by 2.

: 2Al + 3Cl<sub>2</sub> → 2AlCl<sub>3</sub>
: Reactants = Products
: 2Al = 2Al
: 6Cl = 6Cl
This equation is now balanced: '''2Al + 3Cl<sub>2</sub> → 2AlCl<sub>3</sub>'''

'''Aluminium + Oxygen → Aluminium Oxide'''
: _Al + _O<sub>2</sub> → _Al<sub>2</sub>O<sub>3</sub>
: Reactants = Products
: 1Al ≠ 2Al
: 2O ≠ 3O
We need the same number of [[Oxygen]] [[atom]]s on both sides. To do that we can find the Lowest Common Multiple. In this case it is 6, so we times the left 2 [[Oxygen]]s by 3 and the right 3 [[Oxygen]]s by 2.

: _Al + 3O<sub>2</sub> → 2Al<sub>2</sub>O<sub>3</sub>
: Reactants = Products
: 1Al ≠ 4Al
: 6O = 6O
The [[Oxygen]] [[atom]]s are now balanced but the [[Aluminium]] [[atom]]s are not. We need 4 times as many [[Aluminium]] [[atom]]s on the left.

: 4Al + 3O<sub>2</sub> → 2Al<sub>2</sub>O<sub>3</sub>
: Reactants = Products
: 4Al = 4Al
: 6O = 6O

This equation is now balanced: '''4Al + 3O<sub>2</sub> → 2Al<sub>2</sub>O<sub>3</sub>'''

==Key Stage 4==
===Meaning===
A [[Balanced Symbol Equation]] is a way to show the [[reactant]]s and [[product]]s in a [[Chemical Reaction|chemical reaction]] using the [[Chemical Formula|chemical formulae]] of the [[reactant]]s and [[product]]s.

===About Balanced Symbol Equations===
: [[Balanced Symbol Equation]]s show the number of [[molecule]]s of each [[chemical]] and the numbers of each [[atom]] within those [[chemical]]s.
: [[Balanced Symbol Equation]]s can be used to find the quantity of each [[chemical]] needed to have a complete [[Chemical Reaction|chemical reaction]].

===Examples===
2H<sub>2</sub>(g) + O<sub>2</sub>(g) → 2H<sub>2</sub>O(l)

C<sub>2</sub>H<sub>4</sub>(g) + Br<sub>2</sub>(g) → C<sub>2</sub>H<sub>4</sub>Br<sub>2</sub>(g)

C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>(s) + 6O<sub>2</sub>(g) → 6H<sub>2</sub>O(l) + 6CO<sub>2</sub>(g)

4Al(s) + 3O<sub>2</sub>(g) → 2Al<sub>2</sub>O<sub>3</sub>(s)

===Calculating the Mass Required for a Complete Reaction===
{| class="wikitable"
| style="height:20px; width:200px; text-align:center;" |
Find the [[mass]] of [[Oxygen]] needed to completely [[oxidise]] all of the [[Magnesium]]:

2Mg + O<sub>2</sub> → 2MgO

48g + x = y

| style="height:20px; width:200px; text-align:center;" |
Find the [[mass]] of [[Oxygen]] needed for the [[Complete Combustion|complete combustion]] of [[Methane]].

CH<sub>4</sub> + 2O<sub>2</sub> → 2H<sub>2</sub>O + CO<sub>2</sub>

32g + x = y

| style="height:20px; width:200px; text-align:center;" |
Find the [[mass]] of [[Hydrochloric Acid]] needed to completely [[Neutralise (Chemistry)|neutralise]] all of the [[Sodium Hydroxide]].

NaOH + HCl → NaCl + H<sub>2</sub>O

20g + x = y + z
|-
| style="height:20px; width:200px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.

M<sub>r</sub> of Mg = 24g

M<sub>r</sub> of O<sub>2</sub> = 16x2

M<sub>r</sub> of O<sub>2</sub> = 32g

| style="height:20px; width:200px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.

M<sub>r</sub> of CH<sub>4</sub> = 16g

M<sub>r</sub> of O<sub>2</sub> = 16x2

M<sub>r</sub> of O<sub>2</sub> = 32g

| style="height:20px; width:200px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.

M<sub>r</sub> of NaOH = 40g

M<sub>r</sub> of HCl = 36.5g

|-
| style="height:20px; width:200px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.

2 [[mole]]s of Mg are needed for every 1 [[mole]] of O<sub>2</sub>

| style="height:20px; width:200px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.

1 [[mole]] of CH<sub>4</sub> is needed for every 2 [[mole]]s of O<sub>2</sub>

| style="height:20px; width:200px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.

1 [[mole]] of HCl are needed for every 1 [[mole]] of NaOH

|-

| style="height:20px; width:200px; text-align:center;" |
Find the number of [[mole]]s supplied of the known [[mass]].

No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s = <math>\frac{48}{24}</math>

No. [[Mole]]s = 2 Mole

Therefore 1 [[mole]] of O<sub>2</sub> is needed.

1 [[mole]] of O<sub>2</sub> = 32g

| style="height:20px; width:200px; text-align:center;" |
Find the number of [[mole]]s supplied of the known [[mass]].

No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s = <math>\frac{32}{16}</math>

No. [[Mole]]s = 2 Mole

Therefore 4 [[mole]]s of O<sub>2</sub> are needed.

4 [[mole]]s of O<sub>2</sub> = 128g

| style="height:20px; width:200px; text-align:center;" |
Find the number of [[mole]]s supplied of the known [[mass]].

No. [[Mole]]s = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s = <math>\frac{20}{40}</math>

No. [[Mole]]s = 0.5 Mole

Therefore 0.5 [[mole]] of HCl is needed.

0.5 [[mole]] of HCl = 18.25g

|}

===Limiting Reactants===
: A [[Limiting Reactant|limiting reactant]] is a [[reactant]] that is not supplied in a large enough quantity for a complete [[Chemical Reaction|reaction]] with the other [[reactant]]s.

2H<sub>2</sub>(g) + O<sub>2</sub>(g) → 2H<sub>2</sub>O(l)

: If there is 4g of [[Hydrogen]] then 32g of [[Oxygen]] is needed for a complete [[Chemical Reaction|reaction]] to occur.
: If there is 4g of [[Hydrogen]] but only 31g of [[Oxygen]] then the [[Oxygen]] is a [[Limiting Reactant|limiting reactant]].
: If there is 3g of [[Hydrogen]] but only 32g of [[Oxygen]] then the [[Hydrogen]] is a [[Limiting Reactant|limiting reactant]].

{| class="wikitable"
|-
| style="height:20px; width:250px; text-align:center;" |If there are are 6g of [[Carbon]] and 15g of [[Oxygen]] in the following [[Chemical Reaction|reaction]]:
C + O<sub>2</sub> → CO<sub>2</sub>
Which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?

| style="height:20px; width:250px; text-align:center;" |If there are are 27g of [[Aluminium]] and 32g of [[Oxygen]] in the following [[Chemical Reaction|reaction]]:
4Al + 3O<sub>2</sub> → 2Al<sub>2</sub>O<sub>3</sub>
Which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?

| style="height:20px; width:250px; text-align:center;" |If there are are 180g of [[Glucose]] and 200g of [[Oxygen]] in the following [[Chemical Reaction|reaction]]:
C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> + 6O<sub>2</sub> → 6H<sub>2</sub>O + 6CO<sub>2</sub>
Which is the [[Limiting Reactant|limiting reactant]] in this [[Chemical Reaction|reaction]]?

|-
| style="height:20px; width:250px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.

M<sub>r</sub> of C = 12g

M<sub>r</sub> of O<sub>2</sub> = 32g

| style="height:20px; width:250px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.

M<sub>r</sub> of NaOH = 40g

M<sub>r</sub> of HCl = 36.5g

| style="height:20px; width:250px; text-align:center;" |
Find the [[Relative Formula Mass]] of the [[reactant]]s.

M<sub>r</sub> of NaOH = 40g

M<sub>r</sub> of HCl = 36.5g
|-
| style="height:20px; width:250px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.

1 [[mole]]s of C is needed for every 1 [[mole]] of O<sub>2</sub>

| style="height:20px; width:250px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.

4 [[mole]]s of Al are needed for every 3 [[mole]]s of O<sub>2</sub>

| style="height:20px; width:250px; text-align:center;" |
State the [[ratio]] of [[mole]]s of each [[chemical]] needed.

1 [[mole]] of C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> is needed for every 6 [[mole]]s of O<sub>2</sub>
|-

| style="height:20px; width:250px; text-align:center;" |
Find the number of [[mole]]s supplied of each [[chemical]].

No. [[Mole]]s of C = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s of C = <math>\frac{6}{12}</math>

No. [[Mole]]s = 0.5 mol

Therefore 0.5 [[mole]]s of O<sub>2</sub> is needed.

No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s of O = <math>\frac{15}{32}</math>

No. [[Mole]]s = 0.46 mol

[[Oxygen]] is the [[Limiting Reactant|limiting reactant]].
1 [[mole]] of O<sub>2</sub> = 32g
| style="height:20px; width:250px; text-align:center;" |
Find the number of [[mole]]s supplied of each [[chemical]].

No. [[Mole]]s of Al = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s of Al = <math>\frac{27}{27}</math>

No. [[Mole]]s = 1 mol

Therefore 0.75 [[mole]]s of O<sub>2</sub> are needed.

No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s of O = <math>\frac{32}{32}</math>

No. [[Mole]]s = 1 mol

[[Oxygen]] is the [[Limiting Reactant|limiting reactant]].

| style="height:20px; width:250px; text-align:center;" |
Find the number of [[mole]]s supplied of each [[chemical]].

No. [[Mole]]s of C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s of C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> = <math>\frac{180}{180}</math>

No. [[Mole]]s = 1 mol

Therefore 6 [[mole]]s of O<sub>2</sub> are needed.

No. [[Mole]]s of O = <math>\frac{Mass}{M_r}</math>

No. [[Mole]]s of O = <math>\frac{200}{32}</math>

No. [[Mole]]s = 6.25 mol

[[Glucose]] is the [[Limiting Reactant|limiting reactant]].

|}

===Balancing Equations===
: You may know the [[Chemical Formula|chemical formulae]] of the [[reactant]]s and [[product]]s but not know the number of each [[chemical]] needed. For that you must '''balance the equation'''.
: To '''balance and equation''' you need the same number of each type of [[atom]] on the left as you do on the right.

{| class="wikitable"
|+Magnesium Fluoride + Potassium → Potassium Fluoride + Magnesium
|-
| style="height:20px; width:200px; text-align:center;" |Step 1
| style="height:20px; width:200px; text-align:center;" |Step 2
| style="height:20px; width:200px; text-align:center;" |Step 3
|-
|[[File:MgF2+KStep1.png|center|200px]]
|[[File:MgF2+KStep2.png|center|200px]]
|[[File:MgF2+KStep3.png|center|200px]]
|-
| style="height:20px; width:200px; text-align:left;" |
: _MgF<sub>2</sub> + _K → _KF + _Mg
: Reactants = Products
: 1Mg = 1Mg
: 2F ≠ 1F
: 1K = 1K
We need 2 [[Fluorine]] [[atom]]s in the [[product]]s.
| style="height:20px; width:200px; text-align:left;" |
: _MgF<sub>2</sub> + _K → 2KF + _Mg
: Reactants = Products
: 1Mg = 1Mg
: 2F = 2F
: 1K ≠ 2K
There are now too many [[Potassium]] [[atom]]s in the [[product]]s so we need 2 in the [[reactant]]s.
| style="height:20px; width:200px; text-align:left;" |
: _MgF<sub>2</sub> + 2K → 2KF + _Mg
: Reactants = Products
: 1Mg = 1Mg
: 2F = 2F
: 2K = 2K
This equation is '''balanced''':

: MgF<sub>2</sub> + 2K → 2KF + Mg
|}

====Balancing Equations with Ions====
If the [[Electrical Charge|charge]] on an [[ion]] is known, this information can be used to find the [[Chemical Formula|chemical formula]] of an [[Ionic Compound|ionic compound]] and be used to '''balance''' a '''symbol equation'''.
: [[Group 1]] [[Element]]s all form +1 [[ion]]s; Li<sup>+1</sup>, Na<sup>+1</sup>, K<sup>+1</sup>
: [[Group 2]] [[Element]]s all form +2 [[ion]]s; Be<sup>+2</sup>, Mg<sup>+2</sup>, Ca<sup>+2</sup>
: [[Group 3]] [[Element]]s all form +3 [[ion]]s; Al<sup>+3</sup>
: [[Transition Metal]] [[Element]]s can form different [[ion]]s which are shown by Roman Numerals; 

[[Iron]] can form [[Iron|Fe (II)]] which is [[Iron|Fe<sup>+2</sup>]] or [[Iron|Fe (III)]] is [[Iron|Fe<sup>+3</sup>]],
[[Manganese]] can form [[Manganese|Mn (II)]] which is [[Manganese|Mn<sup>+2</sup>]] or [[Manganese|Mn (IV)]] which is [[Manganese|Mn<sup>+4</sup>]]
: [[Group 5]] [[Element]]s all form -3 [[ion]]s; N<sup>-3</sup>, P<sup>-3</sup>
: [[Group 6]] [[Element]]s all form -2 [[ion]]s; O<sup>-2</sup>, S<sup>-2</sup>
: [[Group 7]] [[Element]]s all form -1 [[ion]]s; F<sup>-1</sup>, Cl<sup>-1</sup>

: Some [[Simple Covalent Molecule|covalent]] [[compound]]s can form [[Negative Ion|negative ions]]; 

[[Carbonate]] forms -2 [[ion]]s CO<Sub>3</sub><sup>-2</sup>

[[Sulphate]] forms -2 [[ion]]s SO<Sub>4</sub><sup>-2</sup>

[[Nitrate]] forms -1 [[ion]]s NO<Sub>3</sub><sup>-1</sup>

{| class="wikitable"
|-
| style="height:20px; width:200px; text-align:center;" |[[Group 1]] with [[Group 5]]
| style="height:20px; width:200px; text-align:center;" |[[Group 3]] with [[Group 6]]
| style="height:20px; width:200px; text-align:center;" |[[Sulphate]]s and [[Carbonate]]s with [[Group 2]]
|-
| style="height:20px; width:200px; text-align:center;" |Since [[Group 1]] forms +1 [[ion]]s and [[Group 5]] forms -3 [[ion]]s there will be 3 [[Group 1]] [[ion]]s for each [[Group 5]] [[ion]].
| style="height:20px; width:200px; text-align:center;" |Since [[Group 3]] forms +3 [[ion]]s and [[Group 6]] forms -2 [[ion]]s there will be 2 [[Group 3]] [[ion]]s for every 3 [[Group 6]] [[ion]]s.
| style="height:20px; width:200px; text-align:center;" |[[Sulphate]] [[ion]]s and [[Carbonate]] [[ion]]s both have a [[Electrical Charge|charge]] of -2 while [[Group 2]] [[ion]]s have a [[Electrical Charge|charge]] of +2 so there will be 1 [[Sulphate]] or [[Carbonate]] for every 1 [[Group 2]] [[ion]].
|-
| style="height:20px; width:200px; text-align:center;" |
Examples

Li<sub>3</sub>N

Na<sub>3</sub>P
| style="height:20px; width:200px; text-align:center;" |
Examples

B<sub>2</sub>S<sub>3</sub>

Al<sub>2</sub>O<sub>3</sub>
| style="height:20px; width:200px; text-align:center;" |
Examples

MgSO<sub>4</sub>

CaCO <sub>3</sub>
|}

===References===
====AQA====
:[https://www.amazon.co.uk/gp/product/0198359381/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0198359381&linkCode=as2&tag=nrjc-21&linkId=47c8d1ae58d8b3a5e2094cd447154558 ''Balanced equations, pages 6-7, 64-67, GCSE Chemistry; Third Edition, Oxford University Press, AQA '']
:[https://www.amazon.co.uk/gp/product/1782945571/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945571&linkCode=as2&tag=nrjc-21&linkId=9e29fad914244909903e5e93f8a01d52 ''Balancing equations, pages 15, 44, GCSE Chemistry; The Revision Guide, CGP, AQA '']
:[https://www.amazon.co.uk/gp/product/1471851354/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851354&linkCode=as2&tag=nrjc-21&linkId=9012a0d354024419214fb3ad5ac44ba0 ''Balancing equations, pages 181, 183-4, 250-1, GCSE Combined Science Trilogy 1, Hodder, AQA '']
:[https://www.amazon.co.uk/gp/product/178294639X/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=178294639X&linkCode=as2&tag=nrjc-21&linkId=51599bb45a2bfaf7c1b6a978b2ca2616 ''Balancing equations, pages 32, 33, 112, 113, GCSE Combined Science Trilogy; Chemistry, CGP, AQA '']
:[https://www.amazon.co.uk/gp/product/1782945962/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945962&linkCode=as2&tag=nrjc-21&linkId=476bb5c8d1dfb5c08ac81b6d4d1c98d8 ''Balancing equations, pages 32, 33, 118, 119, GCSE Chemistry, CGP, AQA '']
:[https://www.amazon.co.uk/gp/product/1471851346/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851346&linkCode=as2&tag=nrjc-21&linkId=3ac654f4b0da781c49c855a1af4c92ea ''Balancing equations, pages 68, 70, 283-4, GCSE Chemistry, Hodder, AQA '']
:[https://www.amazon.co.uk/gp/product/1782945598/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945598&linkCode=as2&tag=nrjc-21&linkId=ad276ad49df77ab4b40ab4fd0fe09705 ''Balancing equations, pages 99, 199, GCSE Combined Science; The Revision Guide, CGP, AQA '']
:[https://www.amazon.co.uk/gp/product/1471851354/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851354&linkCode=as2&tag=nrjc-21&linkId=9012a0d354024419214fb3ad5ac44ba0 ''Balancing equations; deduction of balancing numbers, pages 188-9, GCSE Combined Science Trilogy 1, Hodder, AQA '']
:[https://www.amazon.co.uk/gp/product/1471851346/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851346&linkCode=as2&tag=nrjc-21&linkId=3ac654f4b0da781c49c855a1af4c92ea ''Balancing equations; ionic equations, pages 285-6, GCSE Chemistry, Hodder, AQA '']
:[https://www.amazon.co.uk/gp/product/1471851354/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851354&linkCode=as2&tag=nrjc-21&linkId=9012a0d354024419214fb3ad5ac44ba0 ''Balancing equations; molar ratios, pages 185-6, GCSE Combined Science Trilogy 1, Hodder, AQA '']
:[https://www.amazon.co.uk/gp/product/1471851346/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851346&linkCode=as2&tag=nrjc-21&linkId=3ac654f4b0da781c49c855a1af4c92ea ''Balancing numbers calculations, pages 73-4, GCSE Chemistry, Hodder, AQA '']
====Edexcel====
:[https://www.amazon.co.uk/gp/product/1292120193/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120193&linkCode=as2&tag=nrjc-21&linkId=572df39392fb4200db8391d98ae6314e ''Balanced equations, page 205, GCSE Combined Science, Pearson Edexcel']
:[https://www.amazon.co.uk/gp/product/1292120215/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120215&linkCode=as2&tag=nrjc-21&linkId=8f96ddb76196848bafdb124354e4cf77 ''Balanced equations, page 61, GCSE Chemistry, Pearson, Edexcel']
:[https://www.amazon.co.uk/gp/product/1782945725/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945725&linkCode=as2&tag=nrjc-21&linkId=694be7494de75af3349537d34e13f7f0 ''Balanced equations, pages 12, 31, 32, GCSE Chemistry; The Revision Guide, CGP, Edexcel']
:[https://www.amazon.co.uk/gp/product/1782948147/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782948147&linkCode=as2&tag=nrjc-21&linkId=f63dcd8345f4e49c717b39a228a36c7c ''Balancing equations, pages 23-26, 89, 90, GCSE Chemistry, CGP, Edexcel  '']
:[https://www.amazon.co.uk/gp/product/0198359381/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0198359381&linkCode=as2&tag=nrjc-21&linkId=47c8d1ae58d8b3a5e2094cd447154558 ''Symbol equations, page 6, GCSE Chemistry; Third Edition, Oxford University Press, AQA '']
:[https://www.amazon.co.uk/gp/product/1782945962/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945962&linkCode=as2&tag=nrjc-21&linkId=476bb5c8d1dfb5c08ac81b6d4d1c98d8 ''Symbol equations, pages 31-33, GCSE Chemistry, CGP, AQA '']
:[https://www.amazon.co.uk/gp/product/178294639X/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=178294639X&linkCode=as2&tag=nrjc-21&linkId=51599bb45a2bfaf7c1b6a978b2ca2616 ''Symbol equations, pages 31-33, GCSE Combined Science Trilogy; Chemistry, CGP, AQA '']