Chemical Equations — Writing and Balancing

Why Do We Need Chemical Equations?

Think about it — in the previous section, we described the burning of magnesium like this:

"When magnesium burns in air, it reacts with oxygen to form magnesium oxide."

That's a long sentence. Imagine writing 100 reactions this way! That's why scientists invented a short, universal language — the chemical equation.

Word Equation

The simplest method — just use the names of the substances with an arrow between them:

$$\text{Magnesium} + \text{Oxygen} \rightarrow \text{Magnesium oxide}$$

Here —

• Reactants are written on the LHS of the arrow ($\rightarrow$)
• Products are written on the RHS
• A '+' sign separates substances on each side
• The arrowhead points toward the products

Equation Using Chemical Formulae

Replace the names with chemical formulae — even more concise:

$$Mg + O_2 \rightarrow MgO$$

This is called a skeletal equation — because the number of atoms on the two sides may not yet match.

Key Point: A word equation uses substance names; a chemical equation uses chemical formulae.

[Board Tip] Only the formula version is officially called a 'chemical equation'. A word-only equation is incomplete in board exams.

The Law of Conservation of Mass

This law from Class IX is the foundation of equation balancing —

Law: In a chemical reaction, mass can neither be created nor destroyed.

This means — the number of atoms of each element must be the same before and after the reaction. Not even a single atom can simply 'disappear' or 'appear from nowhere'.

Now Check the Skeletal Equation

$$Mg + O_2 \rightarrow MgO$$

Element	Atoms on LHS	Atoms on RHS
Mg	1	1
O	2	1

Oxygen atoms are not equal — so this equation is unbalanced. We must balance it.

A Properly Balanced Example

$$Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$$

Element	LHS	RHS
Zn	1	1 ✓
H	2	2 ✓
S	1	1 ✓
O	4	4 ✓

Every element matches — this is a balanced equation.

[Looking Ahead — JEE/NEET Level] All of stoichiometry — calculations like how many grams of product form from a given mass of reactant — flows directly from this single law of mass conservation.

The Hit-and-Trial Method — Balancing an Equation

Let us balance this unbalanced equation step by step:

$$Fe + H_2O \rightarrow Fe_3O_4 + H_2 \quad \text{(unbalanced)}$$

Step 1 — Count the atoms

Element	LHS	RHS
Fe	1	3
H	2	2
O	1	4

Step 2 — Start with the most complex compound

Pick the compound with the most atoms — here, $Fe_3O_4$ — and start with the element appearing most in it: oxygen.

Balance oxygen: RHS has 4, LHS has 1 → put coefficient 4 on $H_2O$ in LHS:

$$Fe + 4H_2O \rightarrow Fe_3O_4 + H_2$$

Step 3 — Balance hydrogen

Now LHS has H = $4 \times 2 = 8$, but RHS has H = 2 → put coefficient 4 on $H_2$:

$$Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2$$

Step 4 — Balance iron

LHS has Fe = 1, RHS has Fe = 3 → put coefficient 3 on Fe in LHS:

$$3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2 \quad ✓$$

Step 5 — Verify

Element	LHS	RHS
Fe	3	3 ✓
H	8	8 ✓
O	4	4 ✓

The equation is now balanced. This method is called the hit-and-trial method.

Crucial Rule: Never change the chemical formulae while balancing — only the coefficients in front of substances can be changed. So $H_2O$ must never be written as $H_2O_4$.

State Symbols and Reaction Conditions

To make a balanced equation more informative, we add symbols for the physical states of substances.

State Symbols

Symbol	Meaning	Example
$(s)$	Solid	$Fe(s)$
$(l)$	Liquid	$H_2O(l)$
$(g)$	Gas	$O_2(g)$, $H_2(g)$
$(aq)$	Aqueous (dissolved in water)	$HCl(aq)$, $NaCl(aq)$

When a substance is dissolved in water, we write $(aq)$.

Example — fully balanced with state symbols:

$$3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$$

Here $(g)$ tells us water is being used as steam, not as a liquid.

Reaction Conditions — Above/Below the Arrow

Sometimes the temperature, pressure, or catalyst is also written on the arrow. Example —

$$CO(g) + 2H_2(g) \xrightarrow[\text{340 atm}]{\text{catalyst}} CH_3OH(l)$$

Photosynthesis needs sunlight and chlorophyll —

$$6CO_2(aq) + 12H_2O(l) \xrightarrow[\text{chlorophyll}]{\text{sunlight}} C_6H_{12}O_6(aq) + 6O_2(aq) + 6H_2O(l)$$

Special Symbols Used in Equations

• $\uparrow$ (up arrow): A gas is evolved
• $\downarrow$ (down arrow): A precipitate forms (insoluble solid)
• $\Delta$: Heat is supplied (heating)

[Board Tip] State symbols are required almost every year. Equations without $(s), (l), (g), (aq)$ do not get full marks.

The Complete Workflow — At a Glance

When given a word statement and asked to write a balanced equation, follow these 5 steps:

Step 1 — Write the word equation

Reactants on the left, products on the right — using full names.

Step 2 — Convert to chemical formulae

Write the correct chemical formula for each substance. Some essential formulae —

Substance	Formula	Substance	Formula
Water	$H_2O$	Ammonia	$NH_3$
Oxygen gas	$O_2$	Carbon dioxide	$CO_2$
Hydrogen gas	$H_2$	Sulphuric acid	$H_2SO_4$
Nitrogen gas	$N_2$	Hydrochloric acid	$HCl$
Chlorine gas	$Cl_2$	Sodium hydroxide	$NaOH$

Step 3 — Check the skeletal equation

Count the atoms of each element on the LHS and RHS.

Step 4 — Balance using hit-and-trial

A recommended order — start with the most complex compound, then balance metals → non-metals → hydrogen → oxygen.

Step 5 — Add state symbols

Add $(s), (l), (g), (aq)$ as required.

A Full Example — Step by Step

Statement: "Solutions of barium chloride and sodium sulphate in water react to give an insoluble precipitate of barium sulphate and a solution of sodium chloride."

Step 1 — Word equation: Barium chloride + Sodium sulphate → Barium sulphate + Sodium chloride

Step 2 — With formulae: $BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + NaCl$

Step 3 — Atom count: LHS: Ba=1, Cl=2, Na=2, S=1, O=4 RHS: Ba=1, S=1, O=4, Na=1, Cl=1 → Na and Cl unbalanced

Step 4 — Balance: $BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$

Step 5 — Add state symbols:

$$BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s)\downarrow + 2NaCl(aq)$$

[Board Important] This 5-step format directly applies to many 3- and 5-mark questions.

🧠 Memory Capsule

A one-glance recap to revisit just before the board exam — every key idea about chemical equations and balancing in one place.

1. Skeletal vs. Balanced Equation

• Skeletal: Formulae are correct, but atoms are not equal (e.g., $Mg + O_2 \rightarrow MgO$).
• Balanced: Atoms of every element are equal on both sides (e.g., $2Mg + O_2 \rightarrow 2MgO$).

2. Law of Conservation of Mass

$$\text{Atoms on LHS} = \text{Atoms on RHS}$$ This is the basis of all balancing.

3. Hit-and-Trial — Golden Order

1. Start with the most complex compound (the one with the most atoms).
2. Within that compound, start with the element having the highest atom count.
3. Then follow the order — metal → non-metal → H → O.
4. Only change coefficients — never the formulae.

4. State Symbols — One-Minute Recap

Symbol	Meaning
$(s)$	Solid
$(l)$	Liquid
$(g)$	Gas
$(aq)$	Dissolved in water
$\uparrow$	Gas evolved
$\downarrow$	Precipitate formed
$\Delta$	Heat supplied

5. Formulae You Must Memorise

• $O_2, H_2, N_2, Cl_2$ — diatomic gases
• Acids — $HCl, H_2SO_4, HNO_3$
• Bases — $NaOH, KOH, Ca(OH)_2$
• Oxides — $MgO, CaO, CO_2, Fe_2O_3, Fe_3O_4$

6. Five-Point Final Check on a Balanced Equation

1. Are atoms of each element equal? ✓
2. Are formulae correct? ✓
3. Charges balanced (for ionic equations)? ✓
4. State symbols added? ✓
5. Special conditions ($\Delta$, catalyst) shown on the arrow? ✓

7. Common Pitfalls

• 'Coefficient' goes outside — 2 H₂O; 'subscript' is inside the formula — H₂O.
• Never write $(H_2O)_4$ or $H_2O_4$ — the correct form is $4H_2O$.
• Coefficients should be the smallest whole numbers.

The Bottom Line: Not a single atom missing on either side — only then is the equation balanced.

Solved Examples

Example 1: Balancing a Simple Equation

Balance the following equation: $\quad H_2 + Cl_2 \rightarrow HCl$

Solution:

1. Atom count:

• LHS: H = 2, Cl = 2
• RHS: H = 1, Cl = 1

1. Imbalance: Both elements appear half on the right.
2. Balance: Put coefficient 2 on $HCl$:

$$H_2 + Cl_2 \rightarrow 2HCl$$

4. Recheck: LHS — H=2, Cl=2; RHS — H=2, Cl=2 ✓
5. Final answer: $\quad H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$

Example 2: Balancing the Combustion of Magnesium

Write the balanced chemical equation for the burning of magnesium ribbon in air.

Solution:

1. Word equation: Magnesium + Oxygen → Magnesium oxide
2. Skeletal equation:

$$Mg + O_2 \rightarrow MgO$$

3. Atom count:

• LHS: Mg=1, O=2
• RHS: Mg=1, O=1 → oxygen unbalanced

3. Balance oxygen: put coefficient 2 on $MgO$:

$$Mg + O_2 \rightarrow 2MgO$$

5. Now check Mg: LHS=1, RHS=2 → put coefficient 2 on Mg:

$$2Mg + O_2 \rightarrow 2MgO$$

6. Recheck: Mg=2=2 ✓, O=2=2 ✓
7. Final answer with state symbols:

$$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$$

[Repeatedly asked in Board — typically 2 marks]

Example 3: NCERT Textbook Question (Activity 1.3)

Write the balanced equation for: Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

Solution:

1. With formulae:

$$BaCl_2 + Al_2(SO_4)_3 \rightarrow BaSO_4 + AlCl_3$$

2. Atom count:

• LHS: Ba=1, Cl=2, Al=2, S=3, O=12
• RHS: Ba=1, S=1, O=4, Al=1, Cl=3

2. Start with the complex compound $Al_2(SO_4)_3$:

• Balance Al: put coefficient 2 on $AlCl_3$
• Balance S: put coefficient 3 on $BaSO_4$

$$BaCl_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 + 2AlCl_3$$

4. Now check Ba and Cl:

• Ba: RHS=3 → put coefficient 3 on $BaCl_2$:

$$3BaCl_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 + 2AlCl_3$$

5. Verify: Ba=3=3 ✓, Cl=6=6 ✓, Al=2=2 ✓, S=3=3 ✓, O=12=12 ✓
6. Final answer:

$$3BaCl_2(aq) + Al_2(SO_4)_3(aq) \rightarrow 3BaSO_4(s) + 2AlCl_3(aq)$$

Example 4: Sodium and Water Reaction

Sodium reacts with water to produce sodium hydroxide and hydrogen gas. Write the balanced equation.

Solution:

1. Skeletal:

$$Na + H_2O \rightarrow NaOH + H_2$$

2. Atom count:

• LHS: Na=1, H=2, O=1
• RHS: Na=1, O=1, H=3 → H unbalanced

2. Issue: RHS has 3 H, LHS has 2 H — odd numbers cause trouble.
3. Trick: Bring H atoms to a common multiple (4) on both sides.

• Multiply $H_2O$ by 2 on LHS:

$$Na + 2H_2O \rightarrow NaOH + H_2$$

5. Now: LHS — Na=1, H=4, O=2; RHS — Na=1, H=3, O=1 → still unbalanced
6. Multiply NaOH by 2:

$$Na + 2H_2O \rightarrow 2NaOH + H_2$$

7. Now: LHS — Na=1, H=4, O=2; RHS — Na=2, H=4, O=2 → Na unbalanced
8. Multiply Na by 2:

$$2Na + 2H_2O \rightarrow 2NaOH + H_2$$

9. Verify: Na=2=2 ✓, H=4=4 ✓, O=2=2 ✓
10. Final answer:

$$2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)\uparrow$$

Example 5: Conservation of Mass — Find the Unknown

When 5.0 g of calcium carbonate is heated, 2.8 g of calcium oxide and some carbon dioxide are formed. Find the mass of $CO_2$ evolved and verify the law of conservation of mass.

Solution:

1. Reaction:

$$CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$$

2. Law of conservation of mass:

$$\text{Mass of reactants} = \text{Mass of products}$$

3. Given:

• Mass of $CaCO_3$ = 5.0 g
• Mass of $CaO$ = 2.8 g

3. Calculation:

$$\text{Mass of CO}_2 = 5.0 - 2.8 = 2.2 \text{ g}$$

5. Verification:

$$\underbrace{5.0 \text{ g}}_{\text{LHS}} = \underbrace{2.8 + 2.2 = 5.0 \text{ g}}_{\text{RHS}} \quad ✓$$

6. Answer: Mass of $CO_2$ evolved = 2.2 g, and the law of conservation of mass is verified.

[Board Important — 3-mark question]

Example 6: Numerical — Formation of Ammonia

Write the balanced equation for the formation of ammonia ($NH_3$). If 14 g of nitrogen reacts completely, find the mass of ammonia formed. (Atomic masses: N=14, H=1)

Solution:

1. Skeletal equation:

$$N_2 + H_2 \rightarrow NH_3$$

2. Balance: Start with N atoms —

• $N_2$ has N = 2 → put coefficient 2 on $NH_3$:
• $2NH_3$ has H = 6 → put coefficient 3 on $H_2$:

$$N_2 + 3H_2 \rightarrow 2NH_3$$

3. Verify: N=2=2 ✓, H=6=6 ✓
4. Final balanced equation:

$$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$

5. Mass calculation:

• Molar mass of $N_2$ = $2 \times 14 = 28$ g
• Molar mass of $2NH_3$ = $2 \times 17 = 34$ g

5. Ratio: 28 g $N_2$ → 34 g $NH_3$
6. Given — 14 g $N_2$:

$$\text{NH}_3 = \frac{34}{28} \times 14 = 17 \text{ g}$$

8. Answer: 17 g of ammonia is formed.

Example 7: NCERT — Combustion of Hydrogen Sulphide

Hydrogen sulphide gas burns in air to give water and sulphur dioxide. Write the balanced chemical equation.

Solution:

1. Skeletal:

$$H_2S + O_2 \rightarrow H_2O + SO_2$$

2. Atom count:

• LHS: H=2, S=1, O=2
• RHS: H=2, O=3, S=1

2. Oxygen unbalanced — RHS has 3, LHS has 2.
3. Trick: First multiply $H_2S$ by 2 — this neatly cascades:

$$2H_2S + O_2 \rightarrow H_2O + SO_2$$

5. Now: LHS — H=4, S=2, O=2; RHS — H=2, O=3, S=1
6. Multiply $H_2O$ and $SO_2$ by 2:

$$2H_2S + O_2 \rightarrow 2H_2O + 2SO_2$$

7. Now: LHS — H=4, S=2, O=2; RHS — H=4, S=2, O=6 → O unbalanced
8. Multiply $O_2$ by 3:

$$2H_2S + 3O_2 \rightarrow 2H_2O + 2SO_2$$

9. Verify: H=4=4 ✓, S=2=2 ✓, O=6=6 ✓
10. Final answer:

$$2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g)$$

Example 8: Skeletal vs. Balanced

Which of the following is a skeletal equation and which is balanced? Justify. (i) $NH_3 \rightarrow N_2 + H_2$ (ii) $2KClO_3 \rightarrow 2KCl + 3O_2$

Solution:

For (i):

• LHS: N=1, H=3
• RHS: N=2, H=2
• Atoms are not equal — this is a skeletal equation.
• Correct balanced version: $2NH_3 \rightarrow N_2 + 3H_2$

For (ii):

• LHS: K=2, Cl=2, O=6
• RHS: K=2, Cl=2, O=6
• All atoms are equal — this is a balanced equation.

Lesson: Just because an equation uses formulae does not mean it's balanced. Always count atoms to verify.

Example 9: NCERT Textbook Question

Balance the following chemical equation: $\quad HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + H_2O$

Solution:

1. Atom count:

• LHS: H=3, N=1, O=5, Ca=1
• RHS: Ca=1, N=2, O=7, H=2

1. Start with N: LHS=1, RHS=2 → put coefficient 2 on $HNO_3$:

$$2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + H_2O$$

3. Now: LHS — H=4, N=2, O=8, Ca=1; RHS — Ca=1, N=2, O=7, H=2
4. Balance H: put coefficient 2 on $H_2O$:

$$2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O$$

5. Verify:

• LHS: H=4, N=2, O=8, Ca=1
• RHS: Ca=1, N=2, O=8, H=4 ✓

5. Final answer:

$$2HNO_3(aq) + Ca(OH)_2(aq) \rightarrow Ca(NO_3)_2(aq) + 2H_2O(l)$$

This is a neutralisation reaction (acid + base → salt + water).

Example 10: NCERT — Sodium Chloride + Silver Nitrate

When silver nitrate solution is added to sodium chloride solution, an insoluble white precipitate of silver chloride and a solution of sodium nitrate are formed. Write the balanced chemical equation with state symbols.

Solution:

1. Skeletal:

$$NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$$

2. Atom count:

• LHS: Na=1, Cl=1, Ag=1, N=1, O=3
• RHS: Ag=1, Cl=1, Na=1, N=1, O=3

2. Already balanced! ✓
3. Add state symbols:

• $NaCl$ — dissolved in water, $(aq)$
• $AgNO_3$ — dissolved in water, $(aq)$
• $AgCl$ — insoluble precipitate, $(s)\downarrow$
• $NaNO_3$ — stays in solution, $(aq)$

2. Final answer:

$$NaCl(aq) + AgNO_3(aq) \rightarrow AgCl(s)\downarrow + NaNO_3(aq)$$

[Board Important] This is a famous precipitation reaction — the white $AgCl$ precipitate.

Example 11: Conservation of Mass — Statement Verification

A student heated 7 g of iron with sulphur and obtained 11 g of iron sulphide (FeS). Calculate the mass of sulphur used.

Solution:

1. Reaction:

$$Fe(s) + S(s) \xrightarrow{\Delta} FeS(s)$$

2. By conservation of mass:

$$\text{Fe} + \text{S} = \text{FeS}$$

3. Calculation:

$$7 + S = 11 \Rightarrow S = 11 - 7 = 4 \text{ g}$$

4. Answer: Mass of sulphur used = 4 g.
5. Verification: Atomic masses — Fe=56, S=32, FeS=88. Theoretical ratio $\frac{56}{32} = 1.75$, and observed ratio $\frac{7}{4} = 1.75$ ✓

A perfect match — the law of conservation of mass is verified.

Example 12: Balancing Multiple Equations

Balance the following equations: (a) $\quad NaOH + H_2SO_4 \rightarrow Na_2SO_4 + H_2O$ (b) $\quad BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + HCl$

Solution:

For (a):

• Skeletal: LHS — Na=1, O=5, H=3, S=1; RHS — Na=2, S=1, O=5, H=2
• Balance Na: put coefficient 2 on $NaOH$
• Balance H: put coefficient 2 on $H_2O$

$$2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$$

• Verify: Na=2=2 ✓, O=6=6 ✓, H=4=4 ✓, S=1=1 ✓

For (b):

• Skeletal: LHS — Ba=1, Cl=2, H=2, S=1, O=4; RHS — Ba=1, S=1, O=4, H=1, Cl=1
• Both Cl and H halved — put coefficient 2 on $HCl$:

$$BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$$

• Verify: Ba=1=1 ✓, Cl=2=2 ✓, H=2=2 ✓, S=1=1 ✓, O=4=4 ✓

Final answers with state symbols:

$$2NaOH(aq) + H_2SO_4(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)$$

$$BaCl_2(aq) + H_2SO_4(aq) \rightarrow BaSO_4(s)\downarrow + 2HCl(aq)$$

Example 13: Aluminium and Copper Chloride

Aluminium displaces copper from copper chloride solution. Write the balanced equation.

Solution:

1. Word equation: Aluminium + Copper chloride → Aluminium chloride + Copper
2. Skeletal:

$$Al + CuCl_2 \rightarrow AlCl_3 + Cu$$

3. Atom count:

• LHS: Al=1, Cu=1, Cl=2
• RHS: Al=1, Cl=3, Cu=1

3. Cl unbalanced — LHS has 2, RHS has 3. Take LCM = 6:

• put coefficient 3 on $CuCl_2$ (Cl = 6):
• put coefficient 2 on $AlCl_3$ (Cl = 6):

$$Al + 3CuCl_2 \rightarrow 2AlCl_3 + Cu$$

5. Now: LHS — Al=1, Cu=3, Cl=6; RHS — Al=2, Cl=6, Cu=1
6. Balance Al and Cu:

• put coefficient 2 on Al in LHS:
• put coefficient 3 on Cu in RHS:

$$2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu$$

7. Verify: Al=2=2 ✓, Cu=3=3 ✓, Cl=6=6 ✓
8. Final answer:

$$2Al(s) + 3CuCl_2(aq) \rightarrow 2AlCl_3(aq) + 3Cu(s)$$

Example 14: Combustion of Methane

Natural gas (mainly methane) burns to produce carbon dioxide and water. Write the balanced equation.

Solution:

1. Skeletal:

$$CH_4 + O_2 \rightarrow CO_2 + H_2O$$

2. Atom count:

• LHS: C=1, H=4, O=2
• RHS: C=1, O=3, H=2

2. Balance H: put coefficient 2 on $H_2O$:

$$CH_4 + O_2 \rightarrow CO_2 + 2H_2O$$

4. Now: LHS — C=1, H=4, O=2; RHS — C=1, O=4, H=4 → O unbalanced
5. Balance O: put coefficient 2 on $O_2$:

$$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$

6. Verify: C=1=1 ✓, H=4=4 ✓, O=4=4 ✓
7. Final answer with state symbols:

$$CH_4(g) + 2O_2(g) \xrightarrow{\Delta} CO_2(g) + 2H_2O(g) + \text{Energy}$$

This is an exothermic reaction — the same reaction occurs when LPG burns in your kitchen.

Example 15: NCERT — Potassium and Water

Potassium reacts with water to give potassium hydroxide and hydrogen gas. Write the balanced equation.

Solution:

1. Skeletal:

$$K + H_2O \rightarrow KOH + H_2$$

2. Atom count:

• LHS: K=1, H=2, O=1
• RHS: K=1, O=1, H=3 → H unbalanced

2. Bring H to LCM = 4:

• multiply $H_2O$ by 2:
• multiply $KOH$ by 2:

$$K + 2H_2O \rightarrow 2KOH + H_2$$

4. Now: LHS — K=1, H=4, O=2; RHS — K=2, O=2, H=4 → K unbalanced
5. Multiply K by 2:

$$2K + 2H_2O \rightarrow 2KOH + H_2$$

6. Verify: K=2=2 ✓, H=4=4 ✓, O=2=2 ✓
7. Final answer:

$$2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g)\uparrow$$

Caution: Potassium reacts with water so vigorously that the hydrogen evolved often catches fire instantly — it is an extremely reactive metal.

Example 16: A Trickier Organic Equation

Write the balanced equation for the complete combustion of ethane ($C_2H_6$).

Solution:

1. Skeletal:

$$C_2H_6 + O_2 \rightarrow CO_2 + H_2O$$

2. Atom count:

• LHS: C=2, H=6, O=2
• RHS: C=1, O=3, H=2

2. Balance C: put coefficient 2 on $CO_2$.
3. Balance H: put coefficient 3 on $H_2O$:

$$C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O$$

5. Now: LHS — C=2, H=6, O=2; RHS — C=2, O=$4+3=7$, H=6 → O = 2 vs 7 (an odd number!)
6. Trick: Multiply the entire equation by 2 so the oxygen coefficient becomes a whole number:

$$2C_2H_6 + O_2 \rightarrow 4CO_2 + 6H_2O$$

7. Now: LHS — C=4, H=12, O=2; RHS — C=4, O=$8+6=14$, H=12
8. Balance O: put coefficient 7 on $O_2$:

$$2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$$

9. Verify: C=4=4 ✓, H=12=12 ✓, O=14=14 ✓
10. Final answer:

$$2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(l)$$

Lesson: When you hit an odd-vs-even conflict in oxygen, multiply the entire equation by 2. This trick comes up repeatedly in organic combustion equations.