Balancing Equations Worksheet from Everett Community College Tutoring Center, designed for practice in balancing chemical equations.

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 write the coefficients (the numbers in front) that make each equation balanced.

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1) H₃PO₄ + KOH → K₃PO₄ + H₂O
→ Need 3 K on right, so put 3 in front of KOH. That gives 3 OH, which makes 3 H₂O? Wait — left has 3H from acid + 3H from base = 6H total → need 3 H₂O on right.
✔ 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O

2) K + B₂O₃ → K₂O + B
→ Right: 2K, 1B; Left: 1K, 2B → Put 6K on left, 3K₂O on right → then 2B on right → need 2B on left? But B₂O₃ already has 2B. So:
6K + B₂O₃ → 3K₂O + 2B
✔ 6 K + 1 B₂O₃ → 3 K₂O + 2 B

3) HCl + NaOH → NaCl + H₂O
Already balanced!
✔ 1 HCl + 1 NaOH → 1 NaCl + 1 H₂O

4) Na + NaNO₃ → Na₂O + N₂
Left: Na and NaNO₃ → Right: Na₂O and N₂
Try: 10Na + 2NaNO₃ → 6Na₂O + N₂
Check: Na: 10+2=12 → right: 6×2=12 ✔️
N: 2 → right: 2 ✔️
O: 2×3=6 → right: 6 ✔️
✔ 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂

5) C + S₈ → CS₂
S₈ has 8 sulfur → need 4 CS₂ to use 8 S → then need 4 C
✔ 4 C + 1 S₈ → 4 CS₂

6) Na + O₂ → Na₂O
Right: 2Na, 1O → Left: O₂ has 2O → so need 2 Na₂O → that’s 4Na → so 4Na + O₂ → 2Na₂O
✔ 4 Na + 1 O₂ → 2 Na₂O

7) N₂ + O₂ → N₂O₅
Right: 2N, 5O → Left: N₂ has 2N, O₂ has 2O → need 5/2 O₂ → multiply all by 2:
2N₂ + 5O₂ → 2N₂O₅
✔ 2 N₂ + 5 O₂ → 2 N₂O₅

8) H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O
Right: 3Mg, 2PO₄ → so need 2 H₃PO₄ and 3 Mg(OH)₂
Then H: left: 2×3 + 3×2 = 6+6=12 → right: 6 H₂O
✔ 2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O

9) NaOH + H₂CO₃ → Na₂CO₃ + H₂O
Need 2 Na on right → 2 NaOH → then H: 2+2=4 → 2 H₂O
✔ 2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O

10) KOH + HBr → KBr + H₂O
Already balanced.
✔ 1 KOH + 1 HBr → 1 KBr + 1 H₂O

11) Na + O₂ → Na₂O
Same as #6 → 4Na + O₂ → 2Na₂O
✔ 4 Na + 1 O₂ → 2 Na₂O

12) Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O
Right: 2Al, 3CO₃ → so 2 Al(OH)₃ and 3 H₂CO₃
H: left: 2×3 + 3×2 = 6+6=12 → 6 H₂O
✔ 2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O

13) Al + S₈ → Al₂S₃
S₈ → need 8 S → Al₂S₃ has 3 S → LCM of 8 and 3 is 24 → so 3 S₈ → 8 Al₂S₃ → then Al: 16
✔ 16 Al + 3 S₈ → 8 Al₂S₃

14) Cs + N₂ → Cs₃N
Right: 3Cs, 1N → Left: N₂ has 2N → so 2 Cs₃N → 6Cs → 6Cs + N₂ → 2Cs₃N
✔ 6 Cs + 1 N₂ → 2 Cs₃N

15) Mg + Cl₂ → MgCl₂
Already balanced.
✔ 1 Mg + 1 Cl₂ → 1 MgCl₂

16) Rb + RbNO₃ → Rb₂O + N₂
Similar to #4 → try 10Rb + 2RbNO₃ → 6Rb₂O + N₂
Check: Rb: 10+2=12 → 6×2=12 ✔️
N: 2 → 2 ✔️
O: 6 → 6 ✔️
✔ 10 Rb + 2 RbNO₃ → 6 Rb₂O + 1 N₂

17) C₆H₆ + O₂ → CO₂ + H₂O
C: 6 → 6CO₂
H: 6 → 3H₂O
O: right: 6×2 + 3 = 15 → left: O₂ → 15/2 → multiply all by 2:
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
✔ 2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

18) N₂ + H₂ → NH₃
Right: 1N, 3H → Left: N₂, H₂ → need 2NH₃ → then 3H₂
✔ 1 N₂ + 3 H₂ → 2 NH₃

19) C₁₀H₂₂ + O₂ → CO₂ + H₂O
C: 10 → 10CO₂
H: 22 → 11H₂O
O: right: 20 + 11 = 31 → left: 31/2 O₂ → multiply by 2:
2C₁₀H₂₂ + 31O₂ → 20CO₂ + 22H₂O
✔ 2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O

20) Al(OH)₃ + HBr → AlBr₃ + H₂O
Al: 1 → 1
Br: 3 → 3 HBr
H: left: 3 + 3 = 6 → 3 H₂O
✔ 1 Al(OH)₃ + 3 HBr → 1 AlBr₃ + 3 H₂O

21) CH₃CH₂CH₂CH₃ (butane, C₄H₁₀) + O₂ → CO₂ + H₂O
C: 4 → 4CO₂
H: 10 → 5H₂O
O: 8 + 5 = 13 → 13/2 O₂ → ×2:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
✔ 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

22) C₃H₈ + O₂ → CO₂ + H₂O
C: 3 → 3CO₂
H: 8 → 4H₂O
O: 6 + 4 = 10 → 5O₂
✔ 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

23) Li + AlCl₃ → LiCl + Al
Al: 1 → 1
Cl: 3 → 3LiCl → 3Li
✔ 3 Li + 1 AlCl₃ → 3 LiCl + 1 Al

24) C₂H₆ + O₂ → CO₂ + H₂O
C: 2 → 2CO₂
H: 6 → 3H₂O
O: 4 + 3 = 7 → 7/2 O₂ → ×2:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
✔ 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

25) NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + H₂O
Right: 3NH₄ → 3 NH₄OH
H: left: 3×5 + 3 = 15+3=18? Wait — better:
(NH₄)₃PO₄ needs 3 NH₄⁺ → so 3 NH₄OH
H₃PO₄ provides PO₄³⁻
Water: H from 3 NH₄OH (each has 5H? No — NH₄OH is NH₄⁺ and OH⁻ → so 3 NH₄OH has 3×5=15H? Actually, let's count atoms:

Left: 3 NH₄OH → 3N, 15H, 3O
Plus H₃PO₄ → 3H, 1P, 4O
Total left: 3N, 18H, 7O, 1P

Right: (NH₄)₃PO₄ → 3N, 12H, 1P, 4O
Plus H₂O → say x H₂O → 2x H, x O

So: 12 + 2x = 18 → x=3
O: 4 + 3 = 7 ✔️
✔ 3 NH₄OH + 1 H₃PO₄ → 1 (NH₄)₃PO₄ + 3 H₂O

26) Rb + P → Rb₃P
Right: 3Rb, 1P → so 3Rb + P → Rb₃P
✔ 3 Rb + 1 P → 1 Rb₃P

27) CH₄ + O₂ → CO₂ + H₂O
C: 1 → 1CO₂
H: 4 → 2H₂O
O: 2 + 2 = 4 → 2O₂
✔ 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

28) Al(OH)₃ + H₂SO₄ → Al₂(SO₄) + H₂O
Right: 2Al, 3SO₄ → so 2 Al(OH)₃ and 3 H₂SO₄
H: left: 2×3 + 3×2 = 6+6=12 → 6 H₂O
✔ 2 Al(OH)₃ + 3 H₂SO₄ → 1 Al₂(SO₄)₃ + 6 H₂O

29) Na + Cl₂ → NaCl
Need 2NaCl → 2Na + Cl₂
✔ 2 Na + 1 Cl₂ → 2 NaCl

30) Rb + S₈ → Rb₂S
S₈ → 8S → need 4 Rb₂S → 8Rb
✔ 8 Rb + 1 S₈ → 4 Rb₂S

31) H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O
Right: 3Ca, 2PO₄ → so 2 H₃PO₄ and 3 Ca(OH)₂
H: left: 2×3 + 3×2 = 6+6=12 → 6 H₂O
✔ 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O

32) NH₃ + HCl → NH₄Cl
Already balanced.
✔ 1 NH₃ + 1 HCl → 1 NH₄Cl

33) Li + H₂O → LiOH + H₂
Left: Li, 2H, 1O → Right: Li, O, H + H₂ → need 2LiOH and H₂ → then 2Li + 2H₂O → 2LiOH + H₂
✔ 2 Li + 2 H₂O → 2 LiOH + 1 H₂

34) Ca₃(PO₄)₂ + SiO₂ + C → CaSiO₃ + CO + P
This is tricky. Let’s balance step by step.

Assume:
Ca₃(PO₄)₂ → 3Ca, 2P, 8O
SiO₂ → Si, 2O
C → C

Products:
CaSiO₃ → Ca, Si, 3O
CO → C, O
P → P

Try:
Ca₃(PO₄)₂ + 3SiO₂ + 5C → 3CaSiO₃ + 5CO + 2P

Check:
Ca: 3 → 3 ✔️
P: 2 → 2 ✔️
Si: 3 → 3 ✔️
C: 5 → 5 ✔️
O: left: 8 (from phosphate) + 6 (from 3SiO₂) = 14
Right: 3×3=9 (from CaSiO₃) + 5 (from CO) = 14 ✔️
✔ 1 Ca₃(PO₄)₂ + 3 SiO₂ + 5 C → 3 CaSiO₃ + 5 CO + 2 P

35) NH₃ + O₂ → N₂ + H₂O
N: 2 → 2NH₃
H: 6 → 3H₂O
O: 3 → 3/2 O₂ → ×2:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Wait — check:
Left: 4N, 12H, 6O
Right: 4N, 12H, 6O ✔️
✔ 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O

36) FeS₂ + O₂ → Fe₂O₃ + SO₂
Fe: 2 → 2FeS₂
S: 4 → 4SO₂
O: right: 3 + 8 = 11 → left: 11/2 O₂ → ×2:
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
Check:
Fe: 4 → 4 ✔️
S: 8 → 8 ✔️
O: 22 → 6 + 16 = 22 ✔️
✔ 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

37) C + SO₂ → CS₂ + CO
Left: C, S, 2O
Right: CS₂ (1C, 2S), CO (1C, 1O) → total right: 2C, 2S, 1O → not matching.

Try:
3C + 2SO₂ → CS₂ + 2CO
Check:
C: 3 → 1+2=3 ✔️
S: 2 → 2 ✔️
O: 4 → 2 ✔️? Wait — 2SO₂ has 4O → right: 2CO has 2O → missing 2O.

Wrong.

Try:
5C + 2SO₂ → CS₂ + 4CO
C: 5 → 1+4=5 ✔️
S: 2 → 2 ✔️
O: 4 → 4 ✔️
✔ 5 C + 2 SO₂ → 1 CS₂ + 4 CO

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Final Answer:

1) 1, 3, 1, 3
2) 6, 1, 3, 2
3) 1, 1, 1, 1
4) 10, 2, 6, 1
5) 4, 1, 4
6) 4, 1, 2
7) 2, 5, 2
8) 2, 3, 1, 6
9) 2, 1, 1, 2
10) 1, 1, 1, 1
11) 4, 1, 2
12) 2, 3, 1, 6
13) 16, 3, 8
14) 6, 1, 2
15) 1, 1, 1
16) 10, 2, 6, 1
17) 2, 15, 12, 6
18) 1, 3, 2
19) 2, 31, 20, 22
20) 1, 3, 1, 3
21) 2, 13, 8, 10
22) 1, 5, 3, 4
23) 3, 1, 3, 1
24) 2, 7, 4, 6
25) 3, 1, 1, 3
26) 3, 1, 1
27) 1, 2, 1, 2
28) 2, 3, 1, 6
29) 2, 1, 2
30) 8, 1, 4
31) 2, 3, 1, 6
32) 1, 1, 1
33) 2, 2, 2, 1
34) 1, 3, 5, 3, 5, 2
35) 4, 3, 2, 6
36) 4, 11, 2, 8
37) 5, 2, 1, 4