Balancing Equations Worksheet from Everett Community College Tutoring Center.

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 3 H from H₃PO₄ and 3 H from 3 KOH = 6 H total. Right: 3 H₂O has 6 H. Good.
Balanced: 1 H₃PO₄ + 3 KOH → 1 K₃PO₄ + 3 H₂O

2) K + B₂O₃ → K₂O + B
→ Left: 2 B, right: 1 B → put 2 in front of B.
Left: 3 O, right: 1 O in K₂O → need 3 K₂O to get 3 O → then 6 K on right → put 6 K on left.
Balanced: 6 K + 1 B₂O₃ → 3 K₂O + 2 B

3) HCl + NaOH → NaCl + H₂O
Already balanced! All elements match 1:1.
Balanced: 1 HCl + 1 NaOH → 1 NaCl + 1 H₂O

4) Na + NaNO₃ → Na₂O + N₂
This is tricky. Let’s count atoms.
Right: N₂ → needs even N on left. NaNO₃ has 1 N → try 2 NaNO₃ → 2 N → good for N₂.
Now left: 2 Na from NaNO₃ + ? Na metal. Right: Na₂O → let’s say we have x Na₂O → 2x Na and x O.
Left O: 2 NaNO₃ → 6 O → so right must have 6 O → 6 Na₂O → 12 Na on right.
Left: 2 Na from NaNO₃ + y Na → total Na = 2 + y = 12 → y = 10.
Check N: left 2, right 2 (in N₂). Good.
Balanced: 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂

Wait — let me double-check:
Left: Na = 10 + 2 = 12; N = 2; O = 6
Right: Na = 6×2=12; O=6; N=2 → yes!
So: 10 Na + 2 NaNO₃ → 6 Na₂O + 1 N₂

But usually we avoid fractions or large numbers if possible — this seems correct though.

Actually, standard way: sometimes written as 10Na + 2NaNO₃ → 6Na₂O + N₂ — yes.

5) C + S₈ → CS₂
S₈ has 8 S → each CS₂ has 2 S → need 4 CS₂ to use 8 S → then need 4 C.
Balanced: 4 C + 1 S₈ → 4 CS₂

6) Na + O₂ → Na₂O
O₂ has 2 O → Na₂O has 1 O → need 2 Na₂O to use 2 O → then 4 Na on right → put 4 Na on left.
Balanced: 4 Na + 1 O₂ → 2 Na₂O

7) N₂ + O₂ → N₂O₅
N₂O₅ has 2 N and 5 O.
Left: N₂ → 2 N, good. O₂ → need 5 O → but O₂ comes in pairs → least common multiple of 2 and 5 is 10.
So: 2 N₂O₅ → 4 N and 10 O → left: 2 N₂ (for 4 N) and 5 O₂ (for 10 O).
Balanced: 2 N₂ + 5 O₂ → 2 N₂O₅

8) H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O
Mg₃(PO₄)₂ has 3 Mg and 2 PO₄.
So need 2 H₃PO₄ (to give 2 PO₄) and 3 Mg(OH)₂ (to give 3 Mg).
Now H: left: 2×3 = 6 H from acid + 3×2 = 6 H from base = 12 H → right: H₂O → need 6 H₂O (12 H).
O will check out.
Balanced: 2 H₃PO₄ + 3 Mg(OH)₂ → 1 Mg₃(PO₄)₂ + 6 H₂O

9) NaOH + H₂CO₃ → Na₂CO₃ + H₂O
Na₂CO₃ needs 2 Na → so 2 NaOH.
H: left: 2 from NaOH + 2 from H₂CO₃ = 4 H → right: 2 H₂O → 4 H. Good.
Balanced: 2 NaOH + 1 H₂CO₃ → 1 Na₂CO₃ + 2 H₂O

10) KOH + HBr → KBr + H₂O
All 1:1 — already balanced.
Balanced: 1 KOH + 1 HBr → 1 KBr + 1 H₂O

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

12) Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O
Al₂(CO₃)₃ has 2 Al and 3 CO₃.
So need 2 Al(OH)₃ and 3 H₂CO₃.
H: left: 2×3 = 6 H from Al(OH)₃ + 3×2 = 6 H from H₂CO₃ = 12 H → right: 6 H₂O → 12 H.
Balanced: 2 Al(OH)₃ + 3 H₂CO₃ → 1 Al₂(CO₃)₃ + 6 H₂O

13) Al + S₈ → Al₂S₃
S has 8 S. Al₂S₃ has 3 S per formula. LCM of 8 and 3 is 24.
So 3 S₈ → 24 S → need 8 Al₂S₃ (since 8×3=24 S) → then 16 Al on right → put 16 Al on left.
Balanced: 16 Al + 3 S₈ → 8 Al₂S₃

14) Cs + N₂ → Cs₃N
Cs₃N has 3 Cs and 1 N. N₂ has 2 N → so need 2 Cs₃N → 6 Cs and 2 N → left: 6 Cs and 1 N₂.
Balanced: 6 Cs + 1 N₂ → 2 Cs₃N

15) Mg + Cl₂ → MgCl₂
Already balanced — 1 Mg, 2 Cl on each side.
Balanced: 1 Mg + 1 Cl₂ → 1 MgCl₂

16) Rb + RbNO₃ → Rb₂O + N₂
Similar to #4.
RbNO₃ has 1 N → need 2 RbNO₃ for N₂ → 2 N.
O: 2 RbNO₃ → 6 O → Rb₂O has 1 O → need 6 Rb₂O → 12 Rb on right.
Left: 2 Rb from RbNO₃ + x Rb → total Rb = 2 + x = 12 → x = 10.
Balanced: 10 Rb + 2 RbNO₃ → 6 Rb₂O + 1 N₂

17) C₆H₆ + O₂ → CO₂ + H₂O
C₆H₆ has 6 C, 6 H.
Right: 6 CO₂ (for 6 C), 3 H₂O (for 6 H) → now O: 6×2 + 3×1 = 12 + 3 = 15 O → left: O₂ → need 15/2 = 7.5 → multiply all by 2.
So: 2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O
Balanced: 2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

18) N₂ + H₂ → NH₃
NH₃ has 1 N, 3 H. N₂ has 2 N → need 2 NH₃ → 2 N, 6 H → left: 1 N₂, 3 H₂ (6 H).
Balanced: 1 N₂ + 3 H₂ → 2 NH₃

19) C₁₀H₂₂ + O₂ → CO₂ + H₂O
C₁₀H₂₂ → 10 C, 22 H.
Right: 10 CO₂, 11 H₂O (22 H).
O: 10×2 + 11×1 = 20 + 11 = 31 O → left: O₂ → 31/2 → multiply all by 2.
2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O
Balanced: 2 C₁₀H₂₂ + 31 O₂ → 20 CO₂ + 22 H₂O

20) Al(OH)₃ + HBr → AlBr₃ + H₂O
AlBr₃ needs 3 Br → so 3 HBr.
H: left: 3 from Al(OH)₃ + 3 from HBr = 6 H → right: 3 H₂O → 6 H.
Balanced: 1 Al(OH)₃ + 3 HBr → 1 AlBr₃ + 3 H₂O

21) CH₃CH₂CH₂CH₃ + O₂ → CO₂ + H₂O
That’s C₄H₁₀.
C₄H₁₀ → 4 C, 10 H.
Right: 4 CO₂, 5 H₂O (10 H).
O: 4×2 + 5×1 = 8 + 5 = 13 O → left: O₂ → 13/2 → multiply by 2.
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Balanced: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

22) C₃H₈ + O₂ → CO₂ + H₂O
C₃H₈ → 3 C, 8 H.
Right: 3 CO₂, 4 H₂O (8 H).
O: 3×2 + 4×1 = 6 + 4 = 10 O → left: 5 O₂.
Balanced: 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

23) Li + AlCl₃ → LiCl + Al
AlCl₃ has 3 Cl → need 3 LiCl → 3 Li on left.
Al: 1 on each side.
Balanced: 3 Li + 1 AlCl₃ → 3 LiCl + 1 Al

24) C₂H₆ + O₂ → CO₂ + H₂O
C₂H₆ → 2 C, 6 H.
Right: 2 CO₂, 3 H₂O (6 H).
O: 2×2 + 3×1 = 4 + 3 = 7 O → left: 7/2 O₂ → multiply by 2.
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
Balanced: 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

25) NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + H₂O
(NH₄)₃PO₄ needs 3 NH₄ → so 3 NH₄OH.
H: left: 3×5 = 15 H from NH₄OH? Wait — NH₄OH is NH₄⁺ and OH, so atoms: N, 5H, O per molecule? Actually, better to think:
NH₄OH contributes NH₄ and OH. H₃PO₄ contributes 3H and PO₄.
Product: (NH₄)₃PO₄ has 3 NH₄ and PO₄. Water from H⁺ and OH.
So: 3 NH₄OH + 1 H₃PO₄ → 1 (NH₄)₃PO₄ + 3 H₂O
Check H: left: 3*(4+1)=15 H from NH₄OH? No — NH₄OH is often written as such, but atom count: N, 5H, O.
Better:
Left: 3 NH₄OH → 3N, 15H, 3O
1 H₃PO₄ → 3H, 1P, 4O
Total left: 3N, 18H, 1P, 7O
Right: (NH₄)₃PO₄ → 3N, 12H, 1P, 4O
3 H₂O → 6H, 3O
Total right: 3N, 18H, 1P, 7O → perfect.
Balanced: 3 NH₄OH + 1 H₃PO₄ → 1 (NH₄)₃PO₄ + 3 H₂O

26) Rb + P → Rb₃P
Rb₃P has 3 Rb, 1 P.
So 3 Rb + 1 P → 1 Rb₃P
Balanced: 3 Rb + 1 P → 1 Rb₃P

27) CH₄ + O₂ → CO₂ + H₂O
CH₄ → 1 C, 4 H.
Right: 1 CO₂, 2 H₂O (4 H).
O: 2 + 2 = 4 O → left: 2 O₂.
Balanced: 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

28) Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O
Al₂(SO₄)₃ has 2 Al, 3 SO₄.
So need 2 Al(OH)₃ and 3 H₂SO₄.
H: left: 2×3 = 6 H from Al(OH)₃ + 3×2 = 6 H from H₂SO₄ = 12 H → right: 6 H₂O → 12 H.
Balanced: 2 Al(OH)₃ + 3 H₂SO₄ → 1 Al₂(SO₄)₃ + 6 H₂O

29) Na + Cl₂ → NaCl
Cl₂ has 2 Cl → need 2 NaCl → 2 Na on left.
Balanced: 2 Na + 1 Cl₂ → 2 NaCl

30) Rb + S₈ → Rb₂S
S₈ has 8 S. Rb₂S has 1 S → need 8 Rb₂S → 16 Rb on right → put 16 Rb on left.
Balanced: 16 Rb + 1 S₈ → 8 Rb₂S

31) H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O
Ca₃(PO₄)₂ has 3 Ca, 2 PO₄.
So need 2 H₃PO₄ and 3 Ca(OH)₂.
H: left: 2×3 = 6 H from acid + 3×2 = 6 H from base = 12 H → right: 6 H₂O → 12 H.
Balanced: 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O

32) NH₃ + HCl → NH₄Cl
Already balanced — 1:1:1.
Balanced: 1 NH₃ + 1 HCl → 1 NH₄Cl

33) Li + H₂O → LiOH + H₂
LiOH has 1 Li, 1 O, 1 H. H₂ has 2 H.
Left: H₂O has 2 H, 1 O.
To balance H: right has LiOH (1H) + H₂ (2H) = 3H? Not matching.
Try 2 Li + 2 H₂O → 2 LiOH + 1 H₂
Left: 2 Li, 4 H, 2 O
Right: 2 Li, 2 O, 2 H from LiOH + 2 H from H₂ = 4 H → good.
Balanced: 2 Li + 2 H₂O → 2 LiOH + 1 H₂

34) Ca₃(PO₄)₂ + SiO₂ + C → CaSiO₃ + CO + P
This is complex. Let’s assign variables or balance step by step.
First, Ca: left 3, right 1 in CaSiO₃ → so 3 CaSiO₃.
P: left 2, right 1 in P → so 2 P.
Si: left 1 in SiO₂, right 3 in 3 CaSiO₃ → so 3 SiO₂.
Now O: left: Ca₃(PO₄)₂ has 8 O, 3 SiO₂ has 6 O → total 14 O.
Right: 3 CaSiO₃ has 9 O, CO has ? O, P has 0.
Also C: left C, right CO.
Let’s say x CO → x O from CO.
Total O right: 9 + x
Set equal: 9 + x = 14 → x = 5 → so 5 CO.
Then C: left 5 C, right 5 C.
Now check all:
Left: Ca₃(PO₄)₂, 3 SiO₂, 5 C
Atoms: Ca=3, P=2, O=8+6=14, Si=3, C=5
Right: 3 CaSiO₃ → Ca=3, Si=3, O=9; 5 CO → C=5, O=5; 2 P → P=2
Total right: Ca=3, Si=3, O=14, C=5, P=2 → perfect.
Balanced: 1 Ca₃(PO₄)₂ + 3 SiO₂ + 5 C → 3 CaSiO₃ + 5 CO + 2 P

35) NH₃ + O₂ → N₂ + H₂O
N: left 1, right 2 → so 2 NH₃.
H: 2×3=6 H → right: 3 H₂O (6 H).
O: right 3 O → left: 3/2 O₂ → multiply by 2.
4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O
Check: N: 4→4, H:12→12, O:6→6 → good.
Balanced: 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O

36) FeS₂ + O₂ → Fe₂O₃ + SO₂
Fe: left 1, right 2 → so 2 FeS₂.
S: 2×2=4 S → right: 4 SO₂.
O: right: Fe₂O₃ has 3 O, 4 SO₂ has 8 O → total 11 O → left: 11/2 O₂ → multiply by 2.
4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂
Check: Fe:4→4, S:8→8, O:22→6+16=22 → good.
Balanced: 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

37) C + SO₂ → CS₂ + CO
CS₂ has 1 C, 2 S. CO has 1 C, 1 O.
Left: SO₂ has 1 S, 2 O.
Need 2 S on right → so 2 SO₂ on left → 2 S, 4 O.
Right: CS₂ uses 2 S, so 1 CS₂. Then CO must use remaining O: 4 O → so 4 CO? But CO has 1 O each → 4 CO → 4 O.
C: right: 1 (from CS₂) + 4 (from CO) = 5 C → left: 5 C.
Balanced: 5 C + 2 SO₂ → 1 CS₂ + 4 CO

Double-check:
Left: C=5, S=2, O=4
Right: CS₂: C=1,S=2; 4 CO: C=4,O=4 → total C=5,S=2,O=4 → perfect.

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Now, compiling all answers with coefficients:

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) 16, 1, 8
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