On this page, I explain the basic of balancing chemical equations. I assume that you understand the basic of chemical formulas and moles, I explain these concepts in another page. If you are looking how to balance redox equations, click in this link.

The outline for this page is:

1. Theory
2. Explained exercises
3. Exercises
4. Comments

Theory

I have created a video in YouTube based on the theory explained here. You can see the video by clicking here.

A chemical equation is an equation used to describe a chemical reaction, stating the chemical species involved and their proportions. For example: for the reaction

Methane + Oxygen   →  Carbon dioxide + water

The equation is

CH₄(g) + 2O₂(g)       →         CO₂(g)+ 2H₂O(g)

The equation indicates that 1 molecule (or mol) of methane gas reacts with 2 molecules (or moles) of oxygen gas to give 1 molecule (or mol) of carbon dioxide gas and 2 molecules (or moles) of water gas (gas at room temperature is a liquid, however, as this reaction is exothermic and generate a considerable amount of heat, the water appears as vapour). Sometimes is required to write in brackets the physical state of the molecules, the state symbols used are:

State symbol	Meaning
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous solution

The compounds on the left are called reactants and on the right products. The number that indicates the number of molecules or mols are called coefficients. In the examples below, the reactants are CH₄ and O₂, the products: CO₂ and H₂O, the coefficients of the O₂ and H₂O is 2 and the coefficient of the CH₄ and CO₂ is 1 (when it is 1, it is usually omitted):

Balancing chemical equations is essential for understanding and predicting the behavior of chemical reactions, and for making quantitative predictions about the amounts of reactants and products involved in a reaction.

I will explain two methods to balance equations: the traditional method (it is easier for most of the cases and it helps to improve your knowledge of chemistry) and the algebraic method (based on solving mathematical equations, I use this method when the traditional method is challenging).

Traditional method

Rules:

1. Write down each formula of each substance involved in the equation.
2. Bear in mind that the final number of each atom in the reactants and in the products should be the same. If you want to sound more technical, you can say that you are applying the law of the conservation of mass.
3. Identify the most complex substance. Try to find an atom (element) or ionic component (like SO₄^2-, NO₃^–, …) that appears in only one reactant and in only one product, if possible, and adjust the coefficients to ensure that the same number of atoms are on both sides of the equation.
4. Continue balancing the equation, leaving the simple atoms at the end.
5. Coefficients can be written as simple fractions (1/2, 2/3, …), if you don’t like fractions, we can multiply the whole equation (i.e. every single molecule in the equation) for an adequate number to make them whole numbers, for example:

Na(s) + H₂O   → NaOH(aq) + ½ H₂(g)

can also be written as

2Na(s) + 2H₂O  → 2NaOH(aq) + H₂(g)

EXAMPLE:

Given the equation

CH₄(g) + O₂(g) → CO₂(g)+ H₂O(g)

I fix CH₄ to 1

1CH₄(g) + O₂(g) → CO₂(g)+ H₂O(g)

As I start with 1 atom of C and 4 atoms of H (no other reactants have C or H), in the products I will finish with 1 atom of C and 4 atoms of H. Let’s go by steps:

1CH₄(g) + O₂(g)  → 1CO₂(g)+ H₂O(g)

1CH4(g) + O₂(g)  → 1CO₂(g)+ 2H2O(g)

The only element left is the O, and we have fixed the molecules of CH₄, CO₂ and H₂O, so the only component remained to fix is the O₂. Let’s work by steps:

1CH₄(g) + O₂(g)  → 1CO₂(g)+ 2H₂O(g)

1CH₄(g) + 2O₂(g) → 1CO₂(g)+ 2H₂O(g)

Algebraic method

(Also known as Bottomley’s method). This method uses algebraic equations. Steps are:

1. Write down each formula of each substance involved in the equation.
2. Assign a letter for each coefficient in the chemical equation
3. Group the letters by atoms or ionic species creating equations making those that appear as reactant in one side (say left) and those of the products on the other (say right)
4. Assign an arbitrary number (typically 1) to a coefficient (letter) 
5. Resolve the equations

Given the equation

CH₄(g) + O₂(g) → CO₂(g)+ H₂O(g)

I assign a letter to each coefficient

aCH₄(g) + bO₂(g) → cCO₂(g)+ dH₂O(g)

I group the letters by elements and creating equations, bearing in mind that the numbers on one side are for the reactants on the other those for products:

• C: as reactant appears once as a (from the CH₄­) and as product appear once as c (from CO₂), i.e.:

C: a = c

• H: as reactant appears 4 times as a (from the CH₄­) and as product appears 2 times as d (from H₂O), i.e.:

H: 4a = 2d

• O: as reactant appears twice b (from the O₂­) and as product appears twice as c (from CO₂) and once as d (1 time from H₂O), i.e.:

O: 2b = 2c + d

Putting all together:

C: a = c

H: 4a = 2d

O: 2b = 2c + d

Now, I fix 1 letter, say a = 1. If a = 1:

(a =) c =>  1 = c

4a = 2d => 4×1 = 2d => d = 4/2 = 2

2b = 2c + d => 2b = 2×1 + 2 = 2 +2 = 4 => b = 4/2 = 2

Replacing a, b, c and d in the original equation:

CH₄(g) + 2O₂(g) → CO₂(g)+ 2H₂O(g)

To ensure that the equation is balanced, we need to check that the number of atoms that appear in the reactants are the same that in the products

Atom	Number in reactants	Number in products
C	1 from CH4 Total: 1	1 from CO2 Total: 1
H	4 from CH4 Total: 4	2 from each H2O: i.e.: 4 Total: 4
O	2 from each O2 Total: 4	2 from each CO2, i.e.: 2 1 from each H2O: i.e.: 2 Total: 4

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Explained exercises

1. TiCl₄ + Mg → Ti  + MgCl₂

Answer: TiCl₄ + 2Mg → Ti  +  2MgCl₂

Steps (traditional):

i. Fixing TiCl₄ in reactants to 1:

1TiCl₄ + Mg → Ti  + MgCl₂

ii.  Adjusting the products to 1 Ti and 4 Cl:

(1)TiCl₄ + Mg  → 1Ti  + 2MgCl₂

iii.  The only element left is Mg from the reactants, as we have 2 Mg from the products, this should be 2 in the reactants:

(1)TiCl₄ + 2Mg → (1)Ti  + 2MgCl₂

i.e: TiCl₄ + 2Mg → Ti  +  2MgCl₂

Steps (algebra):

aTiCl₄ + bMg → cTi  + dMgCl₂

Algebraic equations:

                            Ti: a = c

                            Cl: 4a = 2d

                            Mg: b = d

 Resolution I fix a = 1, then:

                             a = 1

                             c = 1

                             4a = 2d => 4 = 2d => d = 2

                             b = d => b = 2

Result: TiCl₄ + 2Mg → Ti  +  2MgCl₂

Atom	Number in reactants	Number in products
Ti	1	1
Cl	4	2 x 2 = 4
Mg	2	2

2. Na(s) + H₂O → NaOH(aq) + H₂(g)

Answer: Na(s) + H₂O → NaOH(aq) + 1/2H₂(g)

Or: 2Na(s) + 2H₂O → 2NaOH(aq) + H₂(g)

Steps (traditional):

       i.  Fixing Na in reactants to 1:

1Na(s) + H₂O → NaOH(aq) + H₂(g)

       ii. In products, Na only appears in NaOH, then, we fix it to 1

(1)Na(s) + H₂O → (1)NaOH(aq) + H₂(g)

iii. Inspecting what we have so far, we can see that we have left are the atoms of O and the H:

–  the O appears in H₂O (not fixed yet) and NaOH (fixed)

– the H appears in H₂O (not fixed yet), NaOH (fixed) and H₂ (not fixed yet)

therefore, it is easier to use the O and fix the H₂O (if we use the H, we will have 2 to fix: H₂O and H₂). As in the products we have fixed the atoms to O to 1, in the reactants we need to fix it to 1, i.e. the H2O is 1:

(1)Na(s) + (1)H₂O → (1)NaOH(aq) + H₂(g)

iv. The only element left is the H. We have 2 atoms of H in the reactants via H₂O (already fixed) and in the products we have 1 atom of H in the NaOH (already fixed) and 2 atoms of H in the H₂ (not fixed yet). To make ensure that the number of atoms of H in both sides of the equation are equal, each molecule (mol) of H₂ should contribute with 1 atom, so we need ½ molecule (mol) of H₂.

(1)Na(s) + (1)H₂O → (1)NaOH(aq) + (1/2)H₂(g)

v. This is the answer, however, if you don’t want fraction, multiply the whole equation (reactants and products) by 2

2Na(s) + 2H₂O → 2NaOH(aq) + H₂(g)

Steps (Algebra):

i) Inserting the coefficients: aNa(s) + bH₂O → cNaOH(aq) + dH₂(g)

ii) Write the equations:

Na: a = c

H: 2b = c + 2d

O: b = c

iii) Fix 1 coefficient to 1, say a = 1

iv) Solve the system

              If a = 1, then:

As a = c  => c = 1

As b = c => b = 1

We only have d to solve:   2b = c + 2d => 2×1 = 1 + 2xd => d = ½

v) Write the result:

Na(s) + H₂O → NaOH(aq) + 1/2H₂(g)

Or: 2Na(s) + 2H₂O → 2NaOH(aq) + H₂(g)

Counting the atoms (the 1^st number is for the 1^st result given, the number in brackets is the number in the 2^nd results given)

Atom	Number in reactants	Number in products
Na	1(2)	1(2)
H	2(4)	2(4)
O	1(2)	1(2)

3. CH₂OH + O₂ → CO₂ + H2O

Answer: 4CH₂OH + 5O₂ → 4CO₂ +  6H₂O

Steps (traditional):

i. I fix the CH₂OH to 1,

1CH₂OH + O₂ → CO₂ +  3H₂O

ii. and this fix the number of the atoms of C and H on the products (C and H are only provided by the CH₂OH, but the atom of O is also given by O₂). Let’s do by steps:

a) Atoms of C on the products: I only have 1 atom of C from the reactants, then I can only have 1 atom of C on the products, and therefore I only have 1 CO₂:

1CH₂OH + O₂ → 1CO₂ +  H₂O

b) Atoms of H on the products: I only have 3 atoms of H from the reactants, then I can only have 3 atoms of H on the products, all of them in the molecule of H₂O. As each molecule of H₂O has 2 atoms of H, I need 3/2 molecules of H₂O:

1CH₂OH +  O₂ → 1CO₂ +  3/2H₂O

iii. The only atom left is the O and the molecule of O₂ in the reactants.

a) In the products, I have 2 atoms of O from the molecule of CO₂ and 3/2 of atoms from the H₂O, giving me a total of 2 + 3/2 = 7/2

b) In the reactants I have 1 atom of O from the CH₂OH, the ((7/2)-1 = 5/2) needed has to come from the O₂, as each molecule of O₂ contribute with 2 atoms of O, the number has to be divided by 2 ((5/2)/2 = 5/4)

1CH₂OH + 5/4O₂ → 1CO₂ + 3/2H₂O

iv. The equation is solved, to remove the fraction, you can multiply it by 4:

4CH₂OH + 5O₂ → 4CO₂ + 6H₂O

Steps (algebra):

 i) Inset coefficients

aCH₂OH + bO₂ → cCO₂ + dH2O

ii) Equations:

C: a = c

H: 3a = 2d

O: a + 2b = 2c + d

iii) Fixing 1 coefficient, say a = 1

iv) Solve the system

As a = c => c = 1

As 3a = 2d => 3×1 = 2d => d = 3/2

As a + 2b = 2c + d => 1 + 2b = 2×1 + 3/2 => b = 5/4

 v) Write the equation replacing the coefficients by their values:

              CH₂OH + 5/4O₂ → CO₂ + 3/2H₂O

 Or multiplying by 4:

4CH₂OH + 5O₂ → 4CO₂ + 6H₂O

Atom	Number in reactants	Number in products
C	4	4
H	12	12
O	14	14

Exercises

List of exercises. Click in the number of the exercise to go to the exercise.

Number	Question	Answer	Level
1	Mg + O2 →  MgO	2Mg + O2 → 2MgO	Easy
2	Fe + Cl2→ FeCl3	2Fe + 3Cl2 → 2 FeCl3	Easy
3	Ca + H20 → Ca(OH)2 + H2	Ca + 2 H20 → Ca(OH)2 + H2	Easy
4	C7H16 + O2 → CO2 + H2O	C7H16 + 11O2 → 7CO2 + 8H2O	Easy
5	C8H18 + O2 → CO2 + H2O	2C8H18 + 25O2 → 16CO2 + 18H2O	Easy
6	PCl5 + H2O → H3PO4 + HCl	PCl5 + 4H2O → H3PO4 + 5HCl	Medium
7	P4O10 + H2O → H3PO42	P4O10 + 6H2O → 4H3PO4	Medium
8	SiCl4 + H2O → H4SiO4 + HCl	SiCl4 + 4H2O → H4SiO4 + 4HCl	Medium
9	Al + HCl → AlCl3 + H2	2Al + 6HCl → 2AlCl3 + 3H2	Easy
10	Na2CO3 + HCl → NaCl + H2O + CO2	Na2CO3 + 2HCl → 2NaCl + H2O + CO2	Hard
11	Fe2(SO4)3 + KOH → K2SO4 + Fe(OH)3	Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3	Medium
12	Ca3(PO4)2 + SiO2 → P4O10 + CaSiO3	2Ca3(PO4)2 + 6SiO2 → P4O10 + 6CaSiO3	Hard
13	KClO3 → KClO4 + KCl	4KClO3 → 3KClO4 + KCl	Hard
14	Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4	Al2(SO4)3 + 3Ca(OH)2 → 2Al(OH)3 + 3CaSO4	Medium
15	H2SO4 + HI → H2S + I2 + H2O	H2SO4 + 8HI → H2S + 4I2 + 4H2O	Hard
16	Al + O2 → Al2O3	4Al + 3O2 → 2Al2O3	Medium
17	Al(NO3)3 + NaOH → Al(OH)3 + NaNO3	Al(NO3)3 + 3NaOH → Al(OH)3 + 3NaNO3	Medium
18	O2 + CS2 →CO2 + SO2	3O2 + CS2 →CO2 + 2SO2	Medium
19	BaF2 + K3PO4 → Ba3(PO4)2 + KF	3BaF2 + 2K3PO4 → Ba3(PO4)2 + 6KF	Medium
20	H2SO4 +Mg(NO3)2→ MgSO4 + HNO3	H2SO4 +Mg(NO3)2→ MgSO4 + 2HNO3

1. Mg + O₂ →  MgO

Answer: 2Mg + O₂ → 2MgO

Steps:

1Mg + O₂ → MgO

1Mg + O₂ → 1MgO

Mg + ½ O₂ → MgO

2Mg + O₂ → 2 MgO

Atom	Number in reactants	Number in products
Mg	2	2
O	2	2

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2. Fe + Cl₂→ FeCl3

Answer: 2Fe + 3Cl₂ →2FeCl₃

Steps:

Fe + Cl₂ → 2FeCl₃

Fe + 3Cl₂→ 2FeCl₃

2Fe + 3Cl₂ → 2FeCl₃

Atom	Number in reactants	Number in products
Fe	2	2
Cl	6	6

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3. Ca + H₂0 → Ca(OH)₂ + H₂

Answer: Ca + 2 H₂0 → Ca(OH)₂ + H₂

Steps:

1Ca + H₂0 → Ca(OH)₂ + H₂

1Ca + H₂0 → 1Ca(OH)₂ + H₂

1Ca + 2H₂0 → 1Ca(OH)₂ + H₂

1Ca + 2H₂0 → 1Ca(OH)₂ + 1H₂

Ca + 2H₂0 → Ca(OH)₂ + H₂

Atom	Number in reactants	Number in products
Ca	1	1
H	4	4
O	2	2

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4. C₇H₁₆ + O₂ → CO₂ + H₂O

Answer: C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

Steps:

1C₇H₁₆ + O₂ → CO₂ + H₂O

1C₇H₁₆ + O₂ → 7CO₂ + H₂O

1C₇H₁₆ + O₂ → 7CO₂ + 8H₂O

C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

Atoms	Number in reactants	Number in products
C	7	7
H	16	16
O	22	22

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5. C₈H₁₈ + O₂ → CO₂ + H₂O

Answer: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Steps

1C₈H₁₈ + O₂ → CO₂ + H₂O

1C₈H₁₈ + O₂ → 8CO₂ + H₂O

1C₈H₁₈ + O₂ → 8CO₂ + 9H₂O

1C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

1C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Atom	Number in reactants	Number in products
C	16	16
H	32	32
O	50	50

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6. PCl₅ + H₂O → H₃PO₄ + HCl

Answer: PCl₅ + 4H₂O → H₃PO₄ + 5HCl

Steps

1PCl₅ + H₂O → 1H₃PO₄ + HCl

1PCl₅ + H₂O → 1H₃PO₄ + HCl

1PCl₅ + H₂O → 1H₃PO₄ + 5HCl

1PCl₅ + 4H₂O → 1H₃PO₄ + 5HCl

1PCl₅ + 4H₂O → 1H₃PO₄ + 5HCl

PCl₅ + 4H₂O → H₃PO₄ + 5HCl

Atom	Number in reactants	Number in products
P	1	1
Cl	5	5
H	8	8
O	4	4

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7. P₄O₁₀ + H₂O → H₃PO₄

Answer: P₄O₁₀ + 6H₂O → 4H₃PO₄

Steps:

1P₄O₁₀ + H₂O → H₃PO₄

1P₄O₁₀ + H₂O → 4H₃PO₄

1P₄O₁₀ + 6H₂O → 4H₃PO₄

1P₄O₁₀ + 6H₂O → 4H₃PO₄ (step not needed, just checking O atoms)

P₄O₁₀ + 6H₂O → 4H₃PO₄

8. SiCl₄ + H₂O → H₄SiO₄ + HCl

Answer: SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl

Steps:

1SiCl₄ + H₂O → 1H₄SiO₄ + HCl

1SiCl₄ + H₂O → 1H₄SiO₄ + HCl

1SiCl₄ + H₂O → 1H₄SiO₄ + 4HCl

1SiCl₄ + 4H₂O → 1H₄SiO₄ + 4HCl

1SiCl₄ + 4H₂O → 1H₄SiO₄ + 4HCl (step not needed, just checking H atoms)

SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl

Atom	Number in reactants	Number in products
Si	1	1
Cl	4	4
H	8	8
O	4	4

9. Al + HCl → AlCl₃ + H₂

Answer: 2Al + 6HCl → 2AlCl₃ + 3H₂

Steps:

1Al + HCl → 1AlCl₃ + H₂

1Al + HCl → 1AlCl₃ + H₂

1Al + 3HCl → 1AlCl₃ + H₂

1Al + 3HCl → 1AlCl₃ + 3/2H₂

2Al + 6HCl → 2AlCl₃ + 3H₂

2Al + 6HCl → 2AlCl₃ + 3H₂

Atom	Number in reactants	Number in products
Al	2	2
H	6	6
Cl	6	6

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10. Na₂CO₃ + HCl → NaCl + H₂O + CO₂

Answer: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Steps:

1Na₂CO₃ + HCl → 2NaCl + H₂O + CO₂

 1Na₂CO₃ + HCl → 2NaCl + H₂O + CO₂

1Na₂CO₃ + HCl → 2NaCl + H₂O + 1CO₂

1Na₂CO₃ + HCl → 2NaCl + 1H₂O + 1CO₂

1Na₂CO₃ + 2HCl → 2NaCl + 1H₂O + 1CO₂

1Na₂CO₃ + 2HCl → 2NaCl + 1H₂O + 1CO₂ (step not needed, just checking Cl atoms)

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Atom	Number in reactants	Number in products
Na	2	2
C	1	1
O	3	3
H	2	2
Cl	2	2

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11. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃

Answer: Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃

Steps

Note: SO₄^2- and OH^– are ionic components, treated as a whole unit.

1Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃

1Fe₂(SO₄)₃ + KOH → K₂SO₄ + 2Fe(OH)₃

1Fe₂(SO₄)₃ + KOH → 3K₂SO₄ + 2Fe(OH)₃  

1Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃              

1Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃ (step not needed, just checking OH^–)                 

Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)₃

Atom(s)	Number in reactants	Number in products
Fe	2	2
SO4	3	3
K	6	6
OH	6	6

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12. Ca₃(PO₄)₂ + SiO₂ → P₄O₁₀ + CaSiO₃

Answer: 2Ca₃(PO₄)₂ + 6SiO₂ → P₄O₁₀ + 6CaSiO₃

Steps:

1Ca₃(PO₄)₂ + SiO₂ → P₄O₁₀ + CaSiO₃

1Ca₃(PO₄)₂ + SiO₂ → P₄O₁₀ + 3CaSiO₃

1Ca₃(PO₄)₂ + SiO₂ → 1/2P₄O₁₀ + 3CaSiO₃

1Ca₃(PO₄)₂ + 3SiO₂ → 1/2P₄O₁₀ + 3CaSiO₃

1Ca₃(PO₄)₂ + 3SiO₂ → 1/2P₄O₁₀ + 3CaSiO₃              (step not needed, just checking O)

2Ca₃(PO₄)₂ + 6SiO₂ → 1P₄O₁₀ + 6CaSiO₃                 (adjusting coefficients)

2Ca₃(PO₄)₂ + 6SiO₂ → P₄O₁₀ + 6CaSiO₃

Atom	Number in reactants	Number in products
Ca	6	6
P	4	4
O	28	28
Si	6	6

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13. KClO₃ → KClO₄ + KCl

Answer: 4KClO₃ → 3KClO₄ + KCl

This problem is more complex that it seems at the beginning. As K and Cl appears in all the 3 molecules, we start with O, that only appears in 2 of the molecules and we make it equal in both sides of the equation:

4KClO₃ → 3KClO₄ + KCl

The only specie that we have left if KCl. We can start with Cl, or K, but as they are always in the same ratio (1:1), I put both of them together. If you prefer, you can be more systematic and do 1 by 1:

4KClO₃ → 3KClO₄ + 1KCl

4KClO₃ → 3KClO₄ + KCl

If you get stuck in a problem like this, it is worth to try the algebraic method:

aKClO₃ → bKClO₄ + cKCl

Equations:

              K: a = b + c

              Cl: a = b + c

              O: 3a = 4b

Resolution:

              Fix a = 1

              From the O: 3×1 = 4b => b = ¾

              From the K or the Cl: a = b + c => 1 = ¾ + c => c = ¼

Result

              1KClO₃ → ¾ KClO₄ + ¼KCl

              Or multiplying by 4: 4KClO₃ → 3KClO₄ + KCl

Atom	Number in reactants	Number in products
K	4	4
Cl	4	4
O	12	12

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14. Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + CaSO₄

Answer: Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

Steps:

Note: SO₄^2- and OH^– ionic components

1Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + CaSO₄

 1Al₂(SO₄)₃ + Ca(OH)₂ → 2Al(OH)₃ + CaSO₄

1Al₂(SO₄)₃ + Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

1Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

1Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄         (not needed, checking OH^–)

Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

Atom	Number in reactants	Number in products
Al	2	2
SO42-	3	3
Ca	3	3
OH–	6	6

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15. H₂SO₄ + HI → H₂S + I₂ + H₂O

Answer: H₂SO₄ + 8HI → H₂S + 4I₂ + 4H₂O

Steps:

1H₂SO₄ + HI → H₂S + I₂ + H₂O

1H₂SO₄ + HI → 1H₂S + I₂ + H₂O

1H₂SO₄ + HI → 1H₂S + I₂ + 4H₂O

1H₂SO₄ + 8HI → 1H₂S + I₂ + 4H₂O

1H₂SO₄ + 8HI → 1H₂S + 4I₂ + 4H₂O

H₂SO₄ + 8HI → H₂S + 4I₂ + 4H₂O

As this equation is a bit challenging, using the algebra approach might make it easier:

aH₂SO₄ + bHI → cH₂S + dI₂ + eH₂O

Equations:

 H: 2a + b = 2c + 2e

S: a = c

O: 4a = e

I: b = 2d

Resolution:

Fix a = 1

From S: a = c => c = 1

From O: 4a = e => 4×1 = e => e = 4

From H: 2a + b = 2c + 2e => 2×1 + b = 2×1 + 2×4 => 2 + b = 10 => b = 8

From I: b = 2d => 8 = 2d => d = 4

  Answer:

1H₂SO₄ + 8HI → 1H₂S + 4I₂ + 4H₂O

H₂SO₄ + 8HI → H₂S + 4I₂ + 4H₂O

Atom	Number in reactants	Number in products
H	10	10
S	1	1
O	4	4
H	10	10
I	8	8

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16. Al + O₂ → Al₂O₃

Answer: 4Al + 3O₂ → 2Al₂O₃

Steps:

2Al + O₂ → Al₂O₃                  (As I have 2 Al on the Al₂O₃, it seems a good choice to use 2)

2Al + O₂ → 1Al₂O₃

2Al + 3/2O₂ → 1Al₂O₃

4Al + 3O₂ → 2Al₂O₃

4Al + 3O₂ → 2Al₂O₃

Atom	Number in reactants	Number in products
Al	4	4
O	6	6

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17. Al(NO₃)₃ + NaOH → Al(OH)₃ + NaNO₃

Answer: Al(NO₃)₃ + 3NaOH → Al(OH)₃ + 3NaNO₃

Steps:

1Al(NO₃)₃ + NaOH → Al(OH)₃ + NaNO₃

1Al(NO₃)₃ + NaOH → 1Al(OH)₃ + NaNO₃

1Al(NO₃)₃ + NaOH → 1Al(OH)₃ + 3NaNO₃

1Al(NO₃)₃ + 3NaOH → 1Al(OH)₃ + 3NaNO₃

1Al(NO₃)₃ + 3NaOH → 1Al(OH)₃ + 3NaNO₃            (Not needed, checking OH^–)

Al(NO₃)₃ + 3NaOH → Al(OH)₃ + 3NaNO₃

Atom	Number in reactants	Number in products
Al	1	1
NO3–	3	3
OH–	3	3
Na	3	3

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18. O₂ + CS₂ →CO₂ + SO₂

Answer: 3O₂ + CS₂ →CO₂ + 2SO₂

Steps:

O₂ + 1CS₂ →CO₂ + SO₂       (S and C appear 1 reactants and 1 product)

O₂ + 1CS₂ →1CO₂ + SO₂

O₂ + 1CS₂ →1CO₂ + 2SO₂

3O₂ + 1CS₂ →1CO₂ + 2SO₂

3O₂ + CS₂ →CO₂ + 2SO₂

Atom	Number in reactants	Number in products
O	6	6
C	1	1
S	2	2

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19. BaF₂ + K₃PO₄ → Ba₃(PO₄)₂ + KF

Answer: 3BaF₂ + 2K₃PO₄ → Ba₃(PO₄)₂ + 6KF

Steps:

BaF₂ + K₃PO₄ → 1Ba₃(PO₄)₂ + KF

3BaF₂ + K₃PO₄ → 1Ba₃(PO₄)₂ + KF

3BaF₂ + 2K₃PO₄ → 1Ba₃(PO₄)₂ + KF

3BaF₂ + 2K₃PO₄ → 1Ba₃(PO₄)₂ + 6KF

3BaF₂ + 2K₃PO₄ → 1Ba₃(PO₄)₂ + 6KF          (Steps not needed, checking F)

3BaF₂ + 2K₃PO₄ → Ba₃(PO₄)₂ + 6KF

Atom	Number in reactants	Number in products
Ba	3	3
F	6	6
K	6	6
PO43-	2	2

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20. H₂SO₄ + Mg(NO₃)₂→ MgSO₄ + HNO₃

Answer: H₂SO₄ +Mg(NO₃)₂→ MgSO₄ + 2HNO₃

Steps:

H₂SO₄ +1Mg(NO₃)₂→ MgSO₄ + HNO₃

H₂SO₄ +1Mg(NO₃)₂→ 1MgSO₄ + HNO₃

H₂SO₄ +1Mg(NO₃)₂→ 1MgSO₄ + 1HNO₃

1H₂SO₄ +1Mg(NO₃)₂→ 1MgSO₄ + 1HNO₃

H₂SO₄ +Mg(NO₃)₂→ MgSO₄ + 2HNO₃

Atom	Number in reactants	Number in products
H	2	2
SO42-	1	1
Mg	1	1
NO3–	2	2

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