Writing and balancing chemical equations

Count the atoms on both sides

1. Properties of iron

Writing and balancing chemical equations

2. Do the atoms add up?

In a reaction, no atom is created or destroyed. Nitrogen and hydrogen react to give ammonia.

N2 + H2 \to NH3 N X_{2} + H X_{2} NH X_{3}

Pick the right answer

Predict: count the atoms. Is this equation balanced?

A. Yes: N and H appear on both sides
B. No: the atom counts differ
C. We can't tell without the conditions

Explanation

The equation is not balanced: the left side has 2 N atoms and 2 H atoms, while the right side has 1 N atom and 3 H atoms. Atoms are not created or destroyed in a reaction, so coefficients are needed to make the counts equal.

3. Coefficients, not subscripts

Coefficient

The number written before a formula in an equation. It multiplies the whole formula.

In $2Cl₂ 2 C l_{2}$	Meaning
2 (coefficient)	two molecules
2 (subscript)	two Cl atoms in each molecule
total	2 × 2 = 4 Cl atoms

Conditions, like a catalyst or temperature, go over the arrow:

\to[Δ] Δ

4. Worked: water

H2 + O2 \to H2O H X_{2} + O X_{2} H X_{2} O

Not balanced yet

Count: H 2 and 2; O 2 on the left, 1 on the right. O is not balanced.
Put 2 before $H₂O H_{2} O$ : O is now 2 and 2. But H is now 2 on the left and 4 on the right.
Put 2 before $H₂ H_{2}$ : H is 4 and 4.
Check every atom: H 4 = 4, O 2 = 2. Balanced: $2H2 + O2 \to 2H2O 2 H X_{2} + O X_{2} 2 H X_{2} O$ .

Fixing one atom can unbalance another. So always recount at the end.

5. Balance it yourself

Set the coefficients for

N2 + H2 \to NH3 N X_{2} + H X_{2} NH X_{3}

. Make both sides match.

Coefficient of $N2 N X_{2}$ 1

Coefficient of $H2 H X_{2}$ 1

Coefficient of $NH3 NH X_{3}$ 1

N left 2

N right 1

H left 2

H right 3

6. New equations

Balance the equation

Balance the equation.

Al + O2 \to Al2O3 □ Al + □ O X_{2} □ Al X_{2} O X_{3}

Balanced equation

4Al + 3O2 \to 2Al2O3 4 Al + 3 O X_{2} 2 Al X_{2} O X_{3}

Explanation

Al 4 = 4 and O 6 = 6:

4Al + 3O2 \to 2Al2O3 4 Al + 3 O X_{2} 2 Al X_{2} O X_{3}

Need a hint?

Start with O: 2 on the left, 3 on the right. Which number do 2 and 3 both go into?
Make O equal 6 on both sides. Then count Al again.

Pick the right answer

To balance

Mg + O2 \to MgO Mg + O X_{2} MgO

, a student writes

Mg + O2 \to MgO2 Mg + O X_{2} MgO X_{2}

. What is wrong?

A. The subscript was changed, so the product is no longer magnesium oxide
B. Nothing: the atoms are now equal on both sides
C. The arrow should point the other way
D. A coefficient was put before $O₂ O_{2}$

Explanation

Changing the subscript turned the product into

MgO₂ M g O_{2}

, which is a different substance from magnesium oxide. Only coefficients may change when balancing:

2Mg + O2 \to 2MgO 2 Mg + O X_{2} 2 MgO

7. Summary

Reactants → products; conditions are written over the arrow.
Balance by changing coefficients only, never subscripts.
Check: each kind of atom has the same count on both sides.

Back to the hook:

N2 + 3H2 \to 2NH3 N X_{2} + 3 H X_{2} 2 NH X_{3}

has 2 N and 6 H on each side. No atom is lost.

Can you explain why changing

H₂O H_{2} O

H₂O₂ H_{2} O_{2}

is not a way to balance an equation?

Reveal

A subscript is part of the formula: changing

H₂O H_{2} O

H₂O₂ H_{2} O_{2}

turns water into a different substance. Only coefficients may change:

2H2 + O2 \to 2H2O 2 H X_{2} + O X_{2} 2 H X_{2} O

Concepts in this lesson