Balancing Chemical Equations Homework worksheet with 20 equations to balance.

Let’s go through each equation one by one and balance them. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.

We’ll start with #1 and work our way down to #20.

---

1) N₂ + H₂ → NH₃

Left: 2 N, 2 H
Right: 1 N, 3 H

To balance nitrogen: put a 2 in front of NH₃ → now right has 2 N and 6 H
Now left needs 6 H → so put 3 in front of H₂ (since 3 × 2 = 6)

✔ Balanced: 1 N₂ + 3 H₂ → 2 NH₃

---

2) KClO₃ → KCl + O₂

Left: 1 K, 1 Cl, 3 O
Right: 1 K, 1 Cl, 2 O

Oxygen doesn’t match. Least common multiple of 3 and 2 is 6.

So make 6 O on both sides:

Put 2 in front of KClO₃ → gives 6 O
Then put 3 in front of O₂ → also 6 O

Now check K and Cl: 2 K and 2 Cl on left → need 2 KCl on right

✔ Balanced: 2 KClO₃ → 2 KCl + 3 O₂

---

3) NaCl + F₂ → NaF + Cl₂

Left: 1 Na, 1 Cl, 2 F
Right: 1 Na, 1 F, 2 Cl

Fluorine: 2 on left, only 1 on right → put 2 in front of NaF → now 2 F and 2 Na on right
But now Na is 2 on right → need 2 NaCl on left → that gives 2 Cl on left
Cl₂ on right already has 2 Cl → good!

✔ Balanced: 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂

---

4) H₂ + O₂ → H₂O

Left: 2 H, 2 O
Right: 2 H, 1 O

Need 2 O on right → put 2 in front of H₂O → now 4 H and 2 O on right
Left has only 2 H → put 2 in front of H₂ → 4 H

✔ Balanced: 2 H₂ + 1 O₂ → 2 H₂O

---

5) AgNO₃ + MgCl₂ → AgCl + Mg(NO₃)₂

Left: Ag=1, N=1, O=3, Mg=1, Cl=2
Right: Ag=1, Cl=1, Mg=1, N=2, O=6

Mg(NO₃)₂ has 2 NO₃ groups → so we need 2 AgNO₃ on left to get 2 NO₃

Try: 2 AgNO₃ + MgCl₂ → ?

Now left: Ag=2, N=2, O=6, Mg=1, Cl=2

Right: To get 2 Ag, put 2 AgCl → now Cl=2 → matches left
Mg(NO₃)₂ already has Mg=1, N=2, O=6 → perfect

✔ Balanced: 2 AgNO₃ + 1 MgCl₂ → 2 AgCl + 1 Mg(NO₃)₂

---

6) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

Left: Al=1, Br=3, K=2, S=1, O=4
Right: K=1, Br=1, Al=2, S=3, O=12

Al₂(SO₄)₃ has 2 Al and 3 SO₄ → so need 2 AlBr₃ and 3 K₂SO₄ on left

Try: 2 AlBr₃ + 3 K₂SO₄ → ?

Left: Al=2, Br=6, K=6, S=3, O=12

Right: Need 6 KBr (to get 6 K and 6 Br), and 1 Al₂(SO₄)₃ (already has 2 Al, 3 S, 12 O)

✔ Balanced: 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃

---

7) CH₄ + O₂ → CO₂ + H₂O

Left: C=1, H=4, O=2
Right: C=1, H=2, O=3

H: 4 on left → need 2 H₂O on right → now H=4, O=2+1=3? Wait — CO₂ has 2 O, 2 H₂O has 2 O → total 4 O on right

So right: O=4 → left O₂ must be 2 molecules → 2 O₂ = 4 O

Check: Left: C=1, H=4, O=4
Right: C=1, H=4, O=2 (from CO₂) + 2 (from 2 H₂O) = 4 → wait, 2 H₂O has 2 oxygen atoms? No! Each H₂O has 1 O → 2 H₂O has 2 O. CO₂ has 2 O → total 4 O. Yes.

Actually: 2 H₂O has 2 oxygen atoms? Let me correct: H₂O has 1 oxygen per molecule → 2 H₂O = 2 oxygen atoms. CO₂ = 2 oxygen atoms → total 4 oxygen atoms on right.

Left: 2 O₂ = 4 oxygen atoms → yes.

✔ Balanced: 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

---

8) C₃H₈ + O₂ → CO₂ + H₂O

Left: C=3, H=8, O=2
Right: C=1, H=2, O=3

C: need 3 CO₂ → now C=3, O from CO₂ = 6
H: 8 on left → need 4 H₂O → H=8, O from H₂O = 4
Total O on right: 6 + 4 = 10 → so left needs 5 O₂ (since 5×2=10)

✔ Balanced: 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

---

9) C₈H₁₈ + O₂ → CO₂ + H₂O

Left: C=8, H=18, O=2
Right: C=1, H=2, O=3

C: need 8 CO₂ → O from CO₂ = 16
H: 18 → need 9 H₂O → O from H₂O = 9
Total O on right: 16 + 9 = 25 → so left needs 25/2 O₂ → but we can’t have fractions

Multiply entire equation by 2:

2 C₈H₁₈ + ? O₂ → 16 CO₂ + 18 H₂O

O on right: 16×2 + 18×1 = 32 + 18 = 50 → so O₂ needed = 25

✔ Balanced: 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

---

10) FeCl₃ + NaOH → Fe(OH)₃ + NaCl

Left: Fe=1, Cl=3, Na=1, O=1, H=1
Right: Fe=1, O=3, H=3, Na=1, Cl=1

Cl: 3 on left → need 3 NaCl on right → now Na=3 on right → so need 3 NaOH on left

Now left: Na=3, O=3, H=3 → matches right Fe(OH)₃ (which has 3 O and 3 H)

✔ Balanced: 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl

---

11) P₄ + O₂ → P₂O₅

Left: P=4, O=2
Right: P=2, O=5

P: 4 on left → need 2 P₂O₅ on right → now P=4, O=10
Left O₂ must give 10 O → so 5 O₂

✔ Balanced: 1 P₄ + 5 O₂ → 2 P₂O₅

---

12) Na + H₂O → NaOH + H₂

Left: Na=1, H=2, O=1
Right: Na=1, O=1, H=1+2=3? Wait — NaOH has 1 H, H₂ has 2 H → total 3 H on right

Left has 2 H → not balanced.

Try 2 Na + 2 H₂O → 2 NaOH + H₂

Left: Na=2, H=4, O=2
Right: Na=2, O=2, H=2 (from 2 NaOH) + 2 (from H₂) = 4 → yes!

✔ Balanced: 2 Na + 2 H₂O → 2 NaOH + 1 H₂

---

13) Ag₂O → Ag + O₂

Left: Ag=2, O=1
Right: Ag=1, O=2

O: 1 vs 2 → put 2 Ag₂O on left → Ag=4, O=2
Right: need 4 Ag and 1 O₂

✔ Balanced: 2 Ag₂O → 4 Ag + 1 O₂

---

14) S₈ + O₂ → SO₃

Left: S=8, O=2
Right: S=1, O=3

S: 8 on left → need 8 SO₃ on right → O=24
Left O₂ must give 24 O → 12 O₂

✔ Balanced: 1 S₈ + 12 O₂ → 8 SO₃

---

15) CO₂ + H₂O → C₆H₁₂O₆ + O₂

This is photosynthesis.

Left: C=1, H=2, O=3
Right: C=6, H=12, O=6+2=8? Wait — C₆H₁₂O₆ has 6 O, plus O₂ has 2 O → total 8 O

But let’s count properly:

Right: C₆H₁₂O₆ → C=6, H=12, O=6; plus O₂ → say x O₂ → O=6 + 2x

Left: CO₂ and H₂O → suppose a CO₂ + b H₂O

Set up:

C: a = 6
H: 2b = 12 → b=6
O: 2a + b = 6 + 2x → 2(6) + 6 = 12 + 6 = 18 = 6 + 2x → 2x=12 → x=6

So: 6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂

✔ Balanced: 6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂

---

16) K + MgBr₂ → KBr + Mg

Left: K=1, Mg=1, Br=2
Right: K=1, Br=1, Mg=1

Br: 2 on left → need 2 KBr on right → now K=2 on right → so need 2 K on left

✔ Balanced: 2 K + 1 MgBr₂ → 2 KBr + 1 Mg

---

17) KCl + CaCO₃ → CaCl₂ + K₂CO₃

Left: K=1, Cl=1, Ca=1, C=1, O=3
Right: Ca=1, Cl=2, K=2, C=1, O=3

Cl: 1 vs 2 → need 2 KCl on left → now K=2, Cl=2
Right: K₂CO₃ has K=2 → good
CaCl₂ has Cl=2 → good

✔ Balanced: 2 KCl + 1 CaCO₃ → 1 CaCl₂ + 1 K₂CO₃

---

18) HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂

Left: H=1+1=2, N=1, O=3+3=6, Na=1, C=1
Right: Na=1, N=1, O=3+1+2=6, H=2, C=1

Already balanced!

✔ Balanced: 1 HNO₃ + 1 NaHCO₃ → 1 NaNO₃ + 1 H₂O + 1 CO₂

---

19) H₂O + O₂ → H₂O₂

Left: H=2, O=1+2=3
Right: H=2, O=2

Not balanced. Try 2 H₂O + O₂ → 2 H₂O₂

Left: H=4, O=2+2=4
Right: H=4, O=4 → yes!

✔ Balanced: 2 H₂O + 1 O₂ → 2 H₂O₂

---

20) NaBr + CaF₂ → NaF + CaBr₂

Left: Na=1, Br=1, Ca=1, F=2
Right: Na=1, F=1, Ca=1, Br=2

Br: 1 vs 2 → need 2 NaBr on left → now Na=2, Br=2
F: 2 on left → need 2 NaF on right → Na=2 → good
CaBr₂ has Br=2 → good

✔ Balanced: 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂

---

Final Answer:

1) 1 N₂ + 3 H₂ → 2 NH₃
2) 2 KClO₃ → 2 KCl + 3 O₂
3) 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂
4) 2 H₂ + 1 O₂ → 2 H₂O
5) 2 AgNO₃ + 1 MgCl₂ → 2 AgCl + 1 Mg(NO₃)₂
6) 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
7) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
8) 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
9) 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
10) 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl
11) 1 P₄ + 5 O₂ → 2 P₂O₅
12) 2 Na + 2 H₂O → 2 NaOH + 1 H₂
13) 2 Ag₂O → 4 Ag + 1 O₂
14) 1 S₈ + 12 O₂ → 8 SO₃
15) 6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂
16) 2 K + 1 MgBr₂ → 2 KBr + 1 Mg
17) 2 KCl + 1 CaCO₃ → 1 CaCl₂ + 1 K₂CO₃
18) 1 HNO₃ + 1 NaHCO₃ → 1 NaNO₃ + 1 H₂O + 1 CO₂
19) 2 H₂O + 1 O₂ → 2 H₂O₂
20) 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂