Balancing Equations Worksheet for chemistry practice.
Balancing Equations Worksheet with 37 chemical equations to balance, including reactants and products with blank coefficients, from Everett Community College Tutoring Center.
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Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
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Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
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:
---
1) H₃PO₄ + KOH → K₃PO₄ + H₂O
Left: H=3+1=4, P=1, O=4+1=5, K=1
Right: K=3, P=1, O=4+1=5, H=2
Try putting 3 in front of KOH and 3 in front of H₂O:
→ H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
Check:
Left: H = 3 (from acid) + 3×1 (from base) = 6; K=3; O=4+3=7; P=1
Right: K=3; P=1; O=4 + 3×1 = 7; H=3×2=6 ✔
✔ Balanced: 1, 3, 1, 3
---
2) K + B₂O₃ → K₂O + B
Left: K=1, B=2, O=3
Right: K=2, O=1, B=1
Need to get even numbers. Try 6K on left, 3K₂O on right → that gives 6K and 3O on right. But B₂O₃ has 3O → so use 1 B₂O₃? Then need 2B on right.
Try:
6K + 1B₂O₃ → 3K₂O + 2B
Check:
Left: K=6, B=2, O=3
Right: K=6, O=3, B=2 ✔
✔ Balanced: 6, 1, 3, 2
---
3) HCl + NaOH → NaCl + H₂O
Already balanced! One of each.
✔ Balanced: 1, 1, 1, 1
---
4) Na + NaNO₃ → Na₂O + N₂
Left: Na=1+1=2, N=1, O=3
Right: Na=2, O=1, N=2
Need more N on left → try 2NaNO₃ → then N=2, O=6, Na from nitrate=2, plus extra Na.
Set up:
Let’s say: a Na + b NaNO₃ → c Na₂O + d N₂
Balance N: 2d = b → let d=1 → b=2
Then O: 3b = c → 6 = c → c=6
Then Na: a + b = 2c → a + 2 = 12 → a=10
So: 10Na + 2NaNO₃ → 6Na₂O + 1N₂
Check:
Left: Na=10+2=12, N=2, O=6
Right: Na=12, O=6, N=2 ✔
✔ Balanced: 10, 2, 6, 1
---
5) C + S₈ → CS₂
Left: C=1, S=8
Right: C=1, S=2
Need 4 CS₂ to use 8 S → so:
C + S₈ → 4CS₂ → now C=4 on right → need 4C on left
✔ 4C + 1S₈ → 4CS₂
Check: C=4, S=8 both sides ✔
✔ Balanced: 4, 1, 4
---
6) Na + O₂ → Na₂O
Left: Na=1, O=2
Right: Na=2, O=1
Multiply Na₂O by 2 → O=2, Na=4 → so need 4Na on left
✔ 4Na + 1O₂ → 2Na₂O
Check: Na=4, O=2 both sides ✔
✔ Balanced: 4, 1, 2
---
7) N₂ + O₂ → N₂O₅
Left: N=2, O=2
Right: N=2, O=5
LCM of 2 and 5 is 10 → make O=10 on both sides.
So: 2N₂O₅ → needs 4N and 10O → so left: 2N₂ and 5O₂
✔ 2N₂ + 5O₂ → 2N₂O₅
Check: N=4, O=10 both sides ✔
✔ Balanced: 2, 5, 2
---
8) H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O
Right: Mg=3, P=2, O=8+?= wait — Mg₃(PO₄)₂ has 3Mg, 2P, 8O from phosphate, plus water.
Better to count all.
Left: H₃PO₄ has H=3, P=1, O=4; Mg(OH)₂ has Mg=1, O=2, H=2 per unit.
Right: Mg₃(PO₄)₂ → Mg=3, P=2, O=8; H₂O → H=2, O=1 per unit.
To get 2P on right → need 2 H₃PO₄ on left → then H from acid = 6, P=2, O=8
Now need 3Mg on right → so 3 Mg(OH)₂ on left → Mg=3, O=6, H=6
Total left: H=6+6=12, O=8+6=14, P=2, Mg=3
Right: Mg₃(PO₄)₂ → Mg=3, P=2, O=8 → remaining O and H must be in water.
Total O needed on right: 14 → already 8 in phosphate → need 6 more O → so 6 H₂O → which also gives 12 H → matches left!
✔ 2H₃PO₄ + 3Mg(OH)₂ → 1Mg₃(PO₄)₂ + 6H₂O
Check:
Left: H=2×3 + 3×2 = 6+6=12; P=2; O=2×4 + 3×2 = 8+6=14; Mg=3
Right: Mg=3; P=2; O=8 + 6×1=14; H=12 ✔
✔ Balanced: 2, 3, 1, 6
---
9) NaOH + H₂CO₃ → Na₂CO₃ + H₂O
Left: Na=1, O=1+3=4, H=1+2=3, C=1
Right: Na=2, C=1, O=3+1=4, H=2
Need 2Na on left → 2NaOH → then Na=2, O=2, H=2 from base + H₂CO₃ → H=2+2=4, O=2+3=5, C=1
Right: Na₂CO₃ → Na=2, C=1, O=3 → need 2H₂O to get H=4 and O=2 → total O=5 ✔
✔ 2NaOH + 1H₂CO₃ → 1Na₂CO₃ + 2H₂O
Check:
Left: Na=2, O=2+3=5, H=2+2=4, C=1
Right: Na=2, C=1, O=3+2=5, H=4 ✔
✔ Balanced: 2, 1, 1, 2
---
10) KOH + HBr → KBr + H₂O
Already balanced as written.
✔ 1, 1, 1, 1
---
11) Na + O₂ → Na₂O ← Same as #6!
✔ 4, 1, 2
---
12) Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O
Right: Al=2, C=3, O=9 from carbonate + water
Left: Al(OH)₃ → Al=1, O=3, H=3; H₂CO₃ → H=2, C=1, O=3
Need 2Al on left → 2Al(OH)₃ → Al=2, O=6, H=6
Need 3C on left → 3H₂CO₃ → C=3, H=6, O=9
Total left: Al=2, C=3, H=6+6=12, O=6+9=15
Right: Al₂(CO₃)₃ → Al=2, C=3, O=9 → need 6H₂O to get H=12 and O=6 → total O=15 ✔
✔ 2Al(OH)₃ + 3H₂CO₃ → 1Al₂(CO₃)₃ + 6H₂O
Check:
Left: Al=2, C=3, H=6+6=12, O=6+9=15
Right: Al=2, C=3, O=9+6=15, H=12 ✔
✔ Balanced: 2, 3, 1, 6
---
13) Al + S₈ → Al₂S₃
Left: Al=1, S=8
Right: Al=2, S=3
LCM of S: 8 and 3 → 24
So: 3S₈ → 24S → need 8 Al₂S₃ → which needs 16Al
Left: 16Al + 3S₈ → 8Al₂S₃
Check: Al=16, S=24 both sides ✔
✔ Balanced: 16, 3, 8
---
14) Cs + N₂ → Cs₃N
Left: Cs=1, N=2
Right: Cs=3, N=1
Need 2Cs₃N → Cs=6, N=2 → so left: 6Cs + 1N₂
✔ 6Cs + 1N₂ → 2Cs₃N
Check: Cs=6, N=2 both sides ✔
✔ Balanced: 6, 1, 2
---
15) Mg + Cl₂ → MgCl₂
Already balanced.
✔ 1, 1, 1
---
16) Rb + RbNO₃ → Rb₂O + N₂
Similar to #4.
Left: Rb=1+1=2, N=1, O=3
Right: Rb=2, O=1, N=2
Need 2N on right → so 2RbNO₃ on left → N=2, O=6, Rb from nitrate=2
Then Rb₂O: to get O=6 → need 6Rb₂O → Rb=12
Total Rb on left: x + 2 = 12 → x=10
So: 10Rb + 2RbNO₃ → 6Rb₂O + 1N₂
Check:
Left: Rb=10+2=12, N=2, O=6
Right: Rb=12, O=6, N=2 ✔
✔ Balanced: 10, 2, 6, 1
---
17) C₆H₆ + O₂ → CO₂ + H₂O
Combustion.
Left: C=6, H=6, O=?
Right: C=1 per CO₂, H=2 per H₂O
Make C: 6CO₂ → C=6
Make H: 3H₂O → H=6
Now O on right: 6×2 + 3×1 = 12+3=15 → so O₂ must provide 15/2 → multiply entire equation by 2
Original: C₆H₆ + ?O₂ → 6CO₂ + 3H₂O → O needed: 15 → so 15/2 O₂
Multiply all by 2:
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
Check:
Left: C=12, H=12, O=30
Right: C=12, H=12, O=24+6=30 ✔
✔ Balanced: 2, 15, 12, 6
---
18) N₂ + H₂ → NH₃
Classic Haber process.
Left: N=2, H=2
Right: N=1, H=3
Make N: 2NH₃ → N=2, H=6 → so need 3H₂ on left
✔ 1N₂ + 3H₂ → 2NH₃
Check: N=2, H=6 both sides ✔
✔ Balanced: 1, 3, 2
---
19) C₁₀H₂₂ + O₂ → CO₂ + H₂O
Combustion.
Left: C=10, H=22
Right: set 10CO₂ → C=10; 11H₂O → H=22
O on right: 10×2 + 11×1 = 20+11=31 → so O₂ = 31/2 → multiply by 2
2C₁₀H₂₂ + 31O₂ → 20CO₂ + 22H₂O
Check:
Left: C=20, H=44, O=62
Right: C=20, H=44, O=40+22=62 ✔
✔ Balanced: 2, 31, 20, 22
---
20) Al(OH)₃ + HBr → AlBr₃ + H₂O
Left: Al=1, O=3, H=3+1=4, Br=1
Right: Al=1, Br=3, H=2, O=1
Need 3Br on left → 3HBr → then H=3+3=6, Br=3
Right: AlBr₃ → Al=1, Br=3; need 3H₂O to get H=6 and O=3 → matches left O=3
✔ 1Al(OH)₃ + 3HBr → 1AlBr₃ + 3H₂O
Check:
Left: Al=1, O=3, H=3+3=6, Br=3
Right: Al=1, Br=3, H=6, O=3 ✔
✔ Balanced: 1, 3, 1, 3
---
21) CH₃CH₂CH₂CH₃ + O₂ → CO₂ + H₂O
That’s butane: C₄H₁₀
Same as combustion.
C₄H₁₀ + O₂ → 4CO₂ + 5H₂O → O on right: 8+5=13 → O₂=13/2 → ×2
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Check:
Left: C=8, H=20, O=26
Right: C=8, H=20, O=16+10=26 ✔
✔ Balanced: 2, 13, 8, 10
---
22) C₃H₈ + O₂ → CO₂ + H₂O
Propane.
C₃H₈ → 3CO₂ + 4H₂O → O=6+4=10 → O₂=5
✔ 1C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Check: C=3, H=8, O=10 both sides ✔
✔ Balanced: 1, 5, 3, 4
---
23) Li + AlCl₃ → LiCl + Al
Single replacement.
Left: Li=1, Al=1, Cl=3
Right: Li=1, Cl=1, Al=1
Need 3LiCl → so 3Li on left
✔ 3Li + 1AlCl₃ → 3LiCl + 1Al
Check: Li=3, Al=1, Cl=3 both sides ✔
✔ Balanced: 3, 1, 3, 1
---
24) C₂H₆ + O₂ → CO₂ + H₂O
Ethane.
C₂H₆ → 2CO₂ + 3H₂O → O=4+3=7 → O₂=7/2 → ×2
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Check: C=4, H=12, O=14 both sides ✔
✔ Balanced: 2, 7, 4, 6
---
25) NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + H₂O
Left: N=1, H=4+1+3=8? Wait — NH₄OH is NH₄⁺ and OH⁻ → so N=1, H=5, O=1; H₃PO₄: H=3, P=1, O=4
Better: think of it as ammonia water reacting with phosphoric acid.
Product: (NH₄)₃PO₄ → needs 3NH₄ → so 3NH₄OH
Then: 3NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + ?
Left: N=3, H=3×5 + 3 = 15+3=18? Wait — NH₄OH has 5H? Actually, NH₄OH formula is often written as such, but atom count: N=1, H=5, O=1 per unit.
Standard way: 3NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3H₂O
Check atoms:
Left: N=3, H=3×5 + 3 = 15+3=18? No — H₃PO₄ has 3H, each NH₄OH has 5H? That can’t be right.
Actually, NH₄OH is equivalent to NH₃·H₂O, so atoms: N=1, H=7, O=1? This is messy.
Better to treat as ions or standard reaction.
Commonly: 3NH₃ + H₃PO₄ → (NH₄)₃PO₄, but here it's NH₄OH.
Assume NH₄OH provides NH₄⁺ and OH.
So: 3NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3H₂O
Atom count:
Left: N=3, H=3*(4+1) + 3 = 3*5 +3=18? Wait — NH₄OH: if we break it, N=1, H=5 (4 from NH₄, 1 from OH), O=1.
H₃PO₄: H=3, P=1, O=4
Total left: N=3, H=15+3=18, O=3+4=7, P=1
Right: (NH₄)₃PO₄: N=3, H=12, P=1, O=4; 3H₂O: H=6, O=3 → total H=18, O=7 ✔
Yes!
✔ 3NH₄OH + 1H₃PO₄ → 1(NH₄)₃PO₄ + 3H₂O
✔ Balanced: 3, 1, 1, 3
---
26) Rb + P → Rb₃P
Left: Rb=1, P=1
Right: Rb=3, P=1
So need 3Rb on left
✔ 3Rb + 1P → 1Rb₃P
Check: Rb=3, P=1 both sides ✔
✔ Balanced: 3, 1, 1
---
27) CH₄ + O₂ → CO₂ + H₂O
Methane combustion.
CH₄ + 2O₂ → CO₂ + 2H₂O
Check: C=1, H=4, O=4 both sides ✔
✔ Balanced: 1, 2, 1, 2
---
28) Al(OH)₃ + H₂SO₄ → Al₂(SO₄) + H₂O
Left: Al=1, O=3+4=7? Per unit — better scale.
Need 2Al on right → 2Al(OH)₃ → Al=2, O=6, H=6
Need 3SO₄ on right → 3H₂SO₄ → H=6, S=3, O=12
Total left: Al=2, S=3, H=6+6=12, O=6+12=18
Right: Al₂(SO₄)₃ → Al=2, S=3, O=12; need 6H₂O → H=12, O=6 → total O=18 ✔
✔ 2Al(OH)₃ + 3H₂SO₄ → 1Al₂(SO₄)₃ + 6H₂O
Check:
Left: Al=2, S=3, H=6+6=12, O=6+12=18
Right: Al=2, S=3, O=12+6=18, H=12 ✔
✔ Balanced: 2, 3, 1, 6
---
29) Na + Cl₂ → NaCl
Left: Na=1, Cl=2
Right: Na=1, Cl=1
Need 2NaCl → so 2Na on left
✔ 2Na + 1Cl₂ → 2NaCl
Check: Na=2, Cl=2 both sides ✔
✔ Balanced: 2, 1, 2
---
30) Rb + S₈ → Rb₂S
Left: Rb=1, S=8
Right: Rb=2, S=1
Need 8S on right → 8Rb₂S → Rb=16, S=8
Left: 16Rb + 1S₈
✔ 16Rb + 1S₈ → 8Rb₂S
Check: Rb=16, S=8 both sides ✔
✔ Balanced: 16, 1, 8
---
31) H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O
Similar to #8.
Need 2P on right → 2H₃PO₄ → H=6, P=2, O=8
Need 3Ca on right → 3Ca(OH)₂ → Ca=3, O=6, H=6
Total left: H=12, O=14, P=2, Ca=3
Right: Ca₃(PO₄)₂ → Ca=3, P=2, O=8; need 6H₂O → H=12, O=6 → total O=14 ✔
✔ 2H₃PO₄ + 3Ca(OH)₂ → 1Ca₃(PO₄)₂ + 6H₂O
✔ Balanced: 2, 3, 1, 6
---
32) NH₃ + HCl → NH₄Cl
Already balanced.
✔ 1, 1, 1
---
33) Li + H₂O → LiOH + H₂
Left: Li=1, H=2, O=1
Right: Li=1, O=1, H=1+2=3? LiOH has H=1, H₂ has H=2 → total H=3
Not balanced.
Try 2Li + 2H₂O → 2LiOH + H₂
Left: Li=2, H=4, O=2
Right: Li=2, O=2, H=2+2=4 ✔
✔ 2Li + 2H₂O → 2LiOH + 1H₂
✔ Balanced: 2, 2, 2, 1
---
34) Ca₃(PO₄)₂ + SiO₂ + C → CaSiO₃ + CO + P
This is complex. Let’s balance step by step.
Left: Ca=3, P=2, O=8+2=10? Ca₃(PO₄)₂ has O=8, SiO₂ has O=2, C has no O.
Right: CaSiO₃ has Ca=1, Si=1, O=3; CO has C=1, O=1; P is elemental.
Assume products: a CaSiO₃ + b CO + c P
From Ca: 3 = a
From P: 2 = c
From Si: need b SiO₂? Left has SiO₂, say d SiO₂ → then Si=d, so a=d=3
From C: e C → b CO → so e=b
From O: left: from Ca₃(PO₄)₂: 8O, from d SiO₂: 2d O → total O=8+2d
Right: from a CaSiO₃: 3a O, from b CO: b O → total O=3a + b
Set a=3, d=3, c=2
O left: 8 + 2*3 = 14
O right: 3*3 + b = 9 + b → so 9+b=14 → b=5
Then C: e=5
So:
1Ca₃(PO)₂ + 3SiO₂ + 5C → 3CaSiO₃ + 5CO + 2P
Check atoms:
Left: Ca=3, P=2, O=8+6=14, Si=3, C=5
Right: Ca=3, Si=3, O=9+5=14, C=5, P=2 ✔
✔ Balanced: 1, 3, 5, 3, 5, 2
---
35) NH₃ + O₂ → N₂ + H₂O
Left: N=1, H=3, O=2
Right: N=2, H=2, O=1
Make N: 2NH₃ → N=2, H=6
Make H: 3H₂O → H=6, O=3
O on right: 3 → so O₂=3/2 → ×2
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Check:
Left: N=4, H=12, O=6
Right: N=4, H=12, O=6 ✔
✔ Balanced: 4, 3, 2, 6
---
36) FeS₂ + O₂ → Fe₂O₃ + SO₂
Left: Fe=1, S=2, O=2
Right: Fe=2, S=1, O=3+2=5? Fe₂O₃ has O=3, SO₂ has O=2 per S.
Need 2Fe on right → 2FeS₂ → Fe=2, S=4
Then SO₂: 4SO₂ → S=4, O=8
Fe₂O₃: Fe=2, O=3
Total O on right: 8+3=11 → so O₂=11/2 → ×2
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
Check:
Left: Fe=4, S=8, O=22
Right: Fe=4, S=8, O=6+16=22 ✔
✔ Balanced: 4, 11, 2, 8
---
37) C + SO₂ → CS₂ + CO
Left: C=1, S=1, O=2
Right: C=1+1=2, S=2, O=1
Need 2S on right → 2SO₂ on left → S=2, O=4
Then CS₂ → S=2, C=1; CO → C=1, O=1 → total C=2, O=1 → but left O=4 → not matching.
Set: a C + b SO₂ → c CS₂ + d CO
S: b = 2c
O: 2b = d
C: a = c + d
Let c=1 → b=2, d=4, a=1+4=5
So: 5C + 2SO₂ → 1CS₂ + 4CO
Check:
Left: C=5, S=2, O=4
Right: C=1+4=5, S=2, O=4 ✔
✔ Balanced: 5, 2, 1, 4
---
Now, compiling all answers in order:
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
We’ll start with #1:
---
1) H₃PO₄ + KOH → K₃PO₄ + H₂O
Left: H=3+1=4, P=1, O=4+1=5, K=1
Right: K=3, P=1, O=4+1=5, H=2
Try putting 3 in front of KOH and 3 in front of H₂O:
→ H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
Check:
Left: H = 3 (from acid) + 3×1 (from base) = 6; K=3; O=4+3=7; P=1
Right: K=3; P=1; O=4 + 3×1 = 7; H=3×2=6 ✔
✔ Balanced: 1, 3, 1, 3
---
2) K + B₂O₃ → K₂O + B
Left: K=1, B=2, O=3
Right: K=2, O=1, B=1
Need to get even numbers. Try 6K on left, 3K₂O on right → that gives 6K and 3O on right. But B₂O₃ has 3O → so use 1 B₂O₃? Then need 2B on right.
Try:
6K + 1B₂O₃ → 3K₂O + 2B
Check:
Left: K=6, B=2, O=3
Right: K=6, O=3, B=2 ✔
✔ Balanced: 6, 1, 3, 2
---
3) HCl + NaOH → NaCl + H₂O
Already balanced! One of each.
✔ Balanced: 1, 1, 1, 1
---
4) Na + NaNO₃ → Na₂O + N₂
Left: Na=1+1=2, N=1, O=3
Right: Na=2, O=1, N=2
Need more N on left → try 2NaNO₃ → then N=2, O=6, Na from nitrate=2, plus extra Na.
Set up:
Let’s say: a Na + b NaNO₃ → c Na₂O + d N₂
Balance N: 2d = b → let d=1 → b=2
Then O: 3b = c → 6 = c → c=6
Then Na: a + b = 2c → a + 2 = 12 → a=10
So: 10Na + 2NaNO₃ → 6Na₂O + 1N₂
Check:
Left: Na=10+2=12, N=2, O=6
Right: Na=12, O=6, N=2 ✔
✔ Balanced: 10, 2, 6, 1
---
5) C + S₈ → CS₂
Left: C=1, S=8
Right: C=1, S=2
Need 4 CS₂ to use 8 S → so:
C + S₈ → 4CS₂ → now C=4 on right → need 4C on left
✔ 4C + 1S₈ → 4CS₂
Check: C=4, S=8 both sides ✔
✔ Balanced: 4, 1, 4
---
6) Na + O₂ → Na₂O
Left: Na=1, O=2
Right: Na=2, O=1
Multiply Na₂O by 2 → O=2, Na=4 → so need 4Na on left
✔ 4Na + 1O₂ → 2Na₂O
Check: Na=4, O=2 both sides ✔
✔ Balanced: 4, 1, 2
---
7) N₂ + O₂ → N₂O₅
Left: N=2, O=2
Right: N=2, O=5
LCM of 2 and 5 is 10 → make O=10 on both sides.
So: 2N₂O₅ → needs 4N and 10O → so left: 2N₂ and 5O₂
✔ 2N₂ + 5O₂ → 2N₂O₅
Check: N=4, O=10 both sides ✔
✔ Balanced: 2, 5, 2
---
8) H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O
Right: Mg=3, P=2, O=8+?= wait — Mg₃(PO₄)₂ has 3Mg, 2P, 8O from phosphate, plus water.
Better to count all.
Left: H₃PO₄ has H=3, P=1, O=4; Mg(OH)₂ has Mg=1, O=2, H=2 per unit.
Right: Mg₃(PO₄)₂ → Mg=3, P=2, O=8; H₂O → H=2, O=1 per unit.
To get 2P on right → need 2 H₃PO₄ on left → then H from acid = 6, P=2, O=8
Now need 3Mg on right → so 3 Mg(OH)₂ on left → Mg=3, O=6, H=6
Total left: H=6+6=12, O=8+6=14, P=2, Mg=3
Right: Mg₃(PO₄)₂ → Mg=3, P=2, O=8 → remaining O and H must be in water.
Total O needed on right: 14 → already 8 in phosphate → need 6 more O → so 6 H₂O → which also gives 12 H → matches left!
✔ 2H₃PO₄ + 3Mg(OH)₂ → 1Mg₃(PO₄)₂ + 6H₂O
Check:
Left: H=2×3 + 3×2 = 6+6=12; P=2; O=2×4 + 3×2 = 8+6=14; Mg=3
Right: Mg=3; P=2; O=8 + 6×1=14; H=12 ✔
✔ Balanced: 2, 3, 1, 6
---
9) NaOH + H₂CO₃ → Na₂CO₃ + H₂O
Left: Na=1, O=1+3=4, H=1+2=3, C=1
Right: Na=2, C=1, O=3+1=4, H=2
Need 2Na on left → 2NaOH → then Na=2, O=2, H=2 from base + H₂CO₃ → H=2+2=4, O=2+3=5, C=1
Right: Na₂CO₃ → Na=2, C=1, O=3 → need 2H₂O to get H=4 and O=2 → total O=5 ✔
✔ 2NaOH + 1H₂CO₃ → 1Na₂CO₃ + 2H₂O
Check:
Left: Na=2, O=2+3=5, H=2+2=4, C=1
Right: Na=2, C=1, O=3+2=5, H=4 ✔
✔ Balanced: 2, 1, 1, 2
---
10) KOH + HBr → KBr + H₂O
Already balanced as written.
✔ 1, 1, 1, 1
---
11) Na + O₂ → Na₂O ← Same as #6!
✔ 4, 1, 2
---
12) Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O
Right: Al=2, C=3, O=9 from carbonate + water
Left: Al(OH)₃ → Al=1, O=3, H=3; H₂CO₃ → H=2, C=1, O=3
Need 2Al on left → 2Al(OH)₃ → Al=2, O=6, H=6
Need 3C on left → 3H₂CO₃ → C=3, H=6, O=9
Total left: Al=2, C=3, H=6+6=12, O=6+9=15
Right: Al₂(CO₃)₃ → Al=2, C=3, O=9 → need 6H₂O to get H=12 and O=6 → total O=15 ✔
✔ 2Al(OH)₃ + 3H₂CO₃ → 1Al₂(CO₃)₃ + 6H₂O
Check:
Left: Al=2, C=3, H=6+6=12, O=6+9=15
Right: Al=2, C=3, O=9+6=15, H=12 ✔
✔ Balanced: 2, 3, 1, 6
---
13) Al + S₈ → Al₂S₃
Left: Al=1, S=8
Right: Al=2, S=3
LCM of S: 8 and 3 → 24
So: 3S₈ → 24S → need 8 Al₂S₃ → which needs 16Al
Left: 16Al + 3S₈ → 8Al₂S₃
Check: Al=16, S=24 both sides ✔
✔ Balanced: 16, 3, 8
---
14) Cs + N₂ → Cs₃N
Left: Cs=1, N=2
Right: Cs=3, N=1
Need 2Cs₃N → Cs=6, N=2 → so left: 6Cs + 1N₂
✔ 6Cs + 1N₂ → 2Cs₃N
Check: Cs=6, N=2 both sides ✔
✔ Balanced: 6, 1, 2
---
15) Mg + Cl₂ → MgCl₂
Already balanced.
✔ 1, 1, 1
---
16) Rb + RbNO₃ → Rb₂O + N₂
Similar to #4.
Left: Rb=1+1=2, N=1, O=3
Right: Rb=2, O=1, N=2
Need 2N on right → so 2RbNO₃ on left → N=2, O=6, Rb from nitrate=2
Then Rb₂O: to get O=6 → need 6Rb₂O → Rb=12
Total Rb on left: x + 2 = 12 → x=10
So: 10Rb + 2RbNO₃ → 6Rb₂O + 1N₂
Check:
Left: Rb=10+2=12, N=2, O=6
Right: Rb=12, O=6, N=2 ✔
✔ Balanced: 10, 2, 6, 1
---
17) C₆H₆ + O₂ → CO₂ + H₂O
Combustion.
Left: C=6, H=6, O=?
Right: C=1 per CO₂, H=2 per H₂O
Make C: 6CO₂ → C=6
Make H: 3H₂O → H=6
Now O on right: 6×2 + 3×1 = 12+3=15 → so O₂ must provide 15/2 → multiply entire equation by 2
Original: C₆H₆ + ?O₂ → 6CO₂ + 3H₂O → O needed: 15 → so 15/2 O₂
Multiply all by 2:
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
Check:
Left: C=12, H=12, O=30
Right: C=12, H=12, O=24+6=30 ✔
✔ Balanced: 2, 15, 12, 6
---
18) N₂ + H₂ → NH₃
Classic Haber process.
Left: N=2, H=2
Right: N=1, H=3
Make N: 2NH₃ → N=2, H=6 → so need 3H₂ on left
✔ 1N₂ + 3H₂ → 2NH₃
Check: N=2, H=6 both sides ✔
✔ Balanced: 1, 3, 2
---
19) C₁₀H₂₂ + O₂ → CO₂ + H₂O
Combustion.
Left: C=10, H=22
Right: set 10CO₂ → C=10; 11H₂O → H=22
O on right: 10×2 + 11×1 = 20+11=31 → so O₂ = 31/2 → multiply by 2
2C₁₀H₂₂ + 31O₂ → 20CO₂ + 22H₂O
Check:
Left: C=20, H=44, O=62
Right: C=20, H=44, O=40+22=62 ✔
✔ Balanced: 2, 31, 20, 22
---
20) Al(OH)₃ + HBr → AlBr₃ + H₂O
Left: Al=1, O=3, H=3+1=4, Br=1
Right: Al=1, Br=3, H=2, O=1
Need 3Br on left → 3HBr → then H=3+3=6, Br=3
Right: AlBr₃ → Al=1, Br=3; need 3H₂O to get H=6 and O=3 → matches left O=3
✔ 1Al(OH)₃ + 3HBr → 1AlBr₃ + 3H₂O
Check:
Left: Al=1, O=3, H=3+3=6, Br=3
Right: Al=1, Br=3, H=6, O=3 ✔
✔ Balanced: 1, 3, 1, 3
---
21) CH₃CH₂CH₂CH₃ + O₂ → CO₂ + H₂O
That’s butane: C₄H₁₀
Same as combustion.
C₄H₁₀ + O₂ → 4CO₂ + 5H₂O → O on right: 8+5=13 → O₂=13/2 → ×2
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Check:
Left: C=8, H=20, O=26
Right: C=8, H=20, O=16+10=26 ✔
✔ Balanced: 2, 13, 8, 10
---
22) C₃H₈ + O₂ → CO₂ + H₂O
Propane.
C₃H₈ → 3CO₂ + 4H₂O → O=6+4=10 → O₂=5
✔ 1C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Check: C=3, H=8, O=10 both sides ✔
✔ Balanced: 1, 5, 3, 4
---
23) Li + AlCl₃ → LiCl + Al
Single replacement.
Left: Li=1, Al=1, Cl=3
Right: Li=1, Cl=1, Al=1
Need 3LiCl → so 3Li on left
✔ 3Li + 1AlCl₃ → 3LiCl + 1Al
Check: Li=3, Al=1, Cl=3 both sides ✔
✔ Balanced: 3, 1, 3, 1
---
24) C₂H₆ + O₂ → CO₂ + H₂O
Ethane.
C₂H₆ → 2CO₂ + 3H₂O → O=4+3=7 → O₂=7/2 → ×2
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Check: C=4, H=12, O=14 both sides ✔
✔ Balanced: 2, 7, 4, 6
---
25) NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + H₂O
Left: N=1, H=4+1+3=8? Wait — NH₄OH is NH₄⁺ and OH⁻ → so N=1, H=5, O=1; H₃PO₄: H=3, P=1, O=4
Better: think of it as ammonia water reacting with phosphoric acid.
Product: (NH₄)₃PO₄ → needs 3NH₄ → so 3NH₄OH
Then: 3NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + ?
Left: N=3, H=3×5 + 3 = 15+3=18? Wait — NH₄OH has 5H? Actually, NH₄OH formula is often written as such, but atom count: N=1, H=5, O=1 per unit.
Standard way: 3NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3H₂O
Check atoms:
Left: N=3, H=3×5 + 3 = 15+3=18? No — H₃PO₄ has 3H, each NH₄OH has 5H? That can’t be right.
Actually, NH₄OH is equivalent to NH₃·H₂O, so atoms: N=1, H=7, O=1? This is messy.
Better to treat as ions or standard reaction.
Commonly: 3NH₃ + H₃PO₄ → (NH₄)₃PO₄, but here it's NH₄OH.
Assume NH₄OH provides NH₄⁺ and OH.
So: 3NH₄OH + H₃PO₄ → (NH₄)₃PO₄ + 3H₂O
Atom count:
Left: N=3, H=3*(4+1) + 3 = 3*5 +3=18? Wait — NH₄OH: if we break it, N=1, H=5 (4 from NH₄, 1 from OH), O=1.
H₃PO₄: H=3, P=1, O=4
Total left: N=3, H=15+3=18, O=3+4=7, P=1
Right: (NH₄)₃PO₄: N=3, H=12, P=1, O=4; 3H₂O: H=6, O=3 → total H=18, O=7 ✔
Yes!
✔ 3NH₄OH + 1H₃PO₄ → 1(NH₄)₃PO₄ + 3H₂O
✔ Balanced: 3, 1, 1, 3
---
26) Rb + P → Rb₃P
Left: Rb=1, P=1
Right: Rb=3, P=1
So need 3Rb on left
✔ 3Rb + 1P → 1Rb₃P
Check: Rb=3, P=1 both sides ✔
✔ Balanced: 3, 1, 1
---
27) CH₄ + O₂ → CO₂ + H₂O
Methane combustion.
CH₄ + 2O₂ → CO₂ + 2H₂O
Check: C=1, H=4, O=4 both sides ✔
✔ Balanced: 1, 2, 1, 2
---
28) Al(OH)₃ + H₂SO₄ → Al₂(SO₄) + H₂O
Left: Al=1, O=3+4=7? Per unit — better scale.
Need 2Al on right → 2Al(OH)₃ → Al=2, O=6, H=6
Need 3SO₄ on right → 3H₂SO₄ → H=6, S=3, O=12
Total left: Al=2, S=3, H=6+6=12, O=6+12=18
Right: Al₂(SO₄)₃ → Al=2, S=3, O=12; need 6H₂O → H=12, O=6 → total O=18 ✔
✔ 2Al(OH)₃ + 3H₂SO₄ → 1Al₂(SO₄)₃ + 6H₂O
Check:
Left: Al=2, S=3, H=6+6=12, O=6+12=18
Right: Al=2, S=3, O=12+6=18, H=12 ✔
✔ Balanced: 2, 3, 1, 6
---
29) Na + Cl₂ → NaCl
Left: Na=1, Cl=2
Right: Na=1, Cl=1
Need 2NaCl → so 2Na on left
✔ 2Na + 1Cl₂ → 2NaCl
Check: Na=2, Cl=2 both sides ✔
✔ Balanced: 2, 1, 2
---
30) Rb + S₈ → Rb₂S
Left: Rb=1, S=8
Right: Rb=2, S=1
Need 8S on right → 8Rb₂S → Rb=16, S=8
Left: 16Rb + 1S₈
✔ 16Rb + 1S₈ → 8Rb₂S
Check: Rb=16, S=8 both sides ✔
✔ Balanced: 16, 1, 8
---
31) H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O
Similar to #8.
Need 2P on right → 2H₃PO₄ → H=6, P=2, O=8
Need 3Ca on right → 3Ca(OH)₂ → Ca=3, O=6, H=6
Total left: H=12, O=14, P=2, Ca=3
Right: Ca₃(PO₄)₂ → Ca=3, P=2, O=8; need 6H₂O → H=12, O=6 → total O=14 ✔
✔ 2H₃PO₄ + 3Ca(OH)₂ → 1Ca₃(PO₄)₂ + 6H₂O
✔ Balanced: 2, 3, 1, 6
---
32) NH₃ + HCl → NH₄Cl
Already balanced.
✔ 1, 1, 1
---
33) Li + H₂O → LiOH + H₂
Left: Li=1, H=2, O=1
Right: Li=1, O=1, H=1+2=3? LiOH has H=1, H₂ has H=2 → total H=3
Not balanced.
Try 2Li + 2H₂O → 2LiOH + H₂
Left: Li=2, H=4, O=2
Right: Li=2, O=2, H=2+2=4 ✔
✔ 2Li + 2H₂O → 2LiOH + 1H₂
✔ Balanced: 2, 2, 2, 1
---
34) Ca₃(PO₄)₂ + SiO₂ + C → CaSiO₃ + CO + P
This is complex. Let’s balance step by step.
Left: Ca=3, P=2, O=8+2=10? Ca₃(PO₄)₂ has O=8, SiO₂ has O=2, C has no O.
Right: CaSiO₃ has Ca=1, Si=1, O=3; CO has C=1, O=1; P is elemental.
Assume products: a CaSiO₃ + b CO + c P
From Ca: 3 = a
From P: 2 = c
From Si: need b SiO₂? Left has SiO₂, say d SiO₂ → then Si=d, so a=d=3
From C: e C → b CO → so e=b
From O: left: from Ca₃(PO₄)₂: 8O, from d SiO₂: 2d O → total O=8+2d
Right: from a CaSiO₃: 3a O, from b CO: b O → total O=3a + b
Set a=3, d=3, c=2
O left: 8 + 2*3 = 14
O right: 3*3 + b = 9 + b → so 9+b=14 → b=5
Then C: e=5
So:
1Ca₃(PO)₂ + 3SiO₂ + 5C → 3CaSiO₃ + 5CO + 2P
Check atoms:
Left: Ca=3, P=2, O=8+6=14, Si=3, C=5
Right: Ca=3, Si=3, O=9+5=14, C=5, P=2 ✔
✔ Balanced: 1, 3, 5, 3, 5, 2
---
35) NH₃ + O₂ → N₂ + H₂O
Left: N=1, H=3, O=2
Right: N=2, H=2, O=1
Make N: 2NH₃ → N=2, H=6
Make H: 3H₂O → H=6, O=3
O on right: 3 → so O₂=3/2 → ×2
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Check:
Left: N=4, H=12, O=6
Right: N=4, H=12, O=6 ✔
✔ Balanced: 4, 3, 2, 6
---
36) FeS₂ + O₂ → Fe₂O₃ + SO₂
Left: Fe=1, S=2, O=2
Right: Fe=2, S=1, O=3+2=5? Fe₂O₃ has O=3, SO₂ has O=2 per S.
Need 2Fe on right → 2FeS₂ → Fe=2, S=4
Then SO₂: 4SO₂ → S=4, O=8
Fe₂O₃: Fe=2, O=3
Total O on right: 8+3=11 → so O₂=11/2 → ×2
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
Check:
Left: Fe=4, S=8, O=22
Right: Fe=4, S=8, O=6+16=22 ✔
✔ Balanced: 4, 11, 2, 8
---
37) C + SO₂ → CS₂ + CO
Left: C=1, S=1, O=2
Right: C=1+1=2, S=2, O=1
Need 2S on right → 2SO₂ on left → S=2, O=4
Then CS₂ → S=2, C=1; CO → C=1, O=1 → total C=2, O=1 → but left O=4 → not matching.
Set: a C + b SO₂ → c CS₂ + d CO
S: b = 2c
O: 2b = d
C: a = c + d
Let c=1 → b=2, d=4, a=1+4=5
So: 5C + 2SO₂ → 1CS₂ + 4CO
Check:
Left: C=5, S=2, O=4
Right: C=1+4=5, S=2, O=4 ✔
✔ Balanced: 5, 2, 1, 4
---
Now, compiling all answers in order:
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
Parent Tip: Review the logic above to help your child master the concept of balancing equations chemistry worksheet.
