Balancing Equations Worksheet with multiple chemical reactions to balance.

You’ve uploaded a “Balancing Equations Worksheet” with 16 unbalanced chemical equations. I’ll balance each one step by step and explain the reasoning.

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General Rules for Balancing Equations:

- Law of Conservation of Mass: Atoms of each element must be equal on both sides.
- Start with elements that appear in only one compound on each side.
- Save hydrogen and oxygen for last if they appear in multiple compounds.
- Use coefficients (whole numbers) — never change subscripts.

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## ✔ Balanced Equations with Explanations:

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1. H₂ + O₂ → H₂O

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

→ Need to make O atoms equal. Put 2 in front of H₂O → now O=2, but H=4 on right.

→ Put 2 in front of H₂ → now H=4 on left.

✔ Balanced: 2H₂ + O₂ → 2H₂O

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2. H₃PO₄ + KOH → K₃PO₄ + H₂O

This is an acid-base reaction.

Unbalanced:
Left: H=3+1=4, P=1, O=4+1=5, K=1
Right: K=3, P=1, O=4+1=5, H=2

→ Need 3K on left → put 3 before KOH.

Now: Left: H=3+3=6, O=4+3=7, K=3
Right: H=2, O=5 → need to fix H and O.

Put 3 before H₂O → Right: H=6, O=4+3=7 → matches!

✔ Balanced: H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O

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3. K + B₂O₃ → K₂O + B

Unbalanced:
Left: K=1, B=2, O=3
Right: K=2, O=1, B=1

→ Balance B: put 2 before B on right → B=2

→ Balance O: left has 3, right has 1 → put 3 before K₂O → now O=3, K=6

→ Left K=1 → put 6 before K

✔ Balanced: 6K + B₂O₃ → 3K₂O + 2B

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4. HCl + NaOH → NaCl + H₂O

This is a classic neutralization.

Already balanced!

Left: H=1+1=2, Cl=1, Na=1, O=1
Right: Na=1, Cl=1, H=2, O=1

✔ Balanced: HCl + NaOH → NaCl + H₂O

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5. Na + NaNO₃ → Na₂O + N₂

Unbalanced:
Left: Na=1+1=2, N=1, O=3
Right: Na=2, O=1, N=2

→ N: left=1, right=2 → put 2 before NaNO₃ → now N=2, O=6, Na=1+2=3

→ Right: Na=2, O=1 → need to adjust.

Put 3 before Na₂O → now Na=6, O=3 → not matching.

Try:
Let’s assume 2NaNO₃ → gives 2N, 6O, 2Na from nitrate + ? Na metal.

We need 2N on right → so N₂ is fine.

Need 6O on right → so 3Na₂O → gives 6Na and 3O? Wait, 3Na₂O = 6Na + 3O — but we have 6O from 2NaNO₃.

Mismatch.

Actually, this reaction is unusual — let’s balance properly.

Set coefficients:

a Na + b NaNO₃ → c Na₂O + d N₂

Atoms:

Na: a + b = 2c
N: b = 2d
O: 3b = c

From O: c = 3b
From N: d = b/2 → b must be even.

Try b=2 → then c=6, d=1, then Na: a + 2 = 12 → a=10

✔ Balanced: 10Na + 2NaNO₃ → 6Na₂O + N₂

*(Note: This is a redox reaction where sodium reduces nitrate to nitrogen gas.)*

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6. C + S₈ → CS₂

S₈ is 8 sulfur atoms.

Right: CS₂ has 2 S per molecule.

To use all 8 S, need 4 CS₂ → 4C and 8S.

✔ Balanced: 4C + S₈ → 4CS₂

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7. Na + O₂ → Na₂O₂

Sodium peroxide.

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

→ Put 2 before Na

✔ Balanced: 2Na + O₂ → Na₂O₂

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8. N₂ + O₂ → N₂O₅

Dinitrogen pentoxide.

Left: N=2, O=2
Right: N=2, O=5

→ Need 5 O on left → but O₂ gives even numbers → multiply O₂ by 5/2 → not integer.

Multiply entire equation by 2:

2N₂ + 5O₂ → 2N₂O₅

Check: N=4, O=10 on both sides.

✔ Balanced: 2N₂ + 5O₂ → 2N₂O₅

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9. H₃PO₄ + Mg(OH)₂ → Mg₃(PO₄)₂ + H₂O

Acid + base → salt + water.

Mg₃(PO₄)₂ implies 3 Mg and 2 PO₄.

So need 2 H₃PO₄ → gives 2 PO₄.

Need 3 Mg(OH)₂ → gives 3 Mg.

Now H: left = 2×3 + 3×2 = 6+6=12 H
Right: H₂O → need 6 H₂O to get 12 H.

O also balances: left = 2×4 + 3×2 = 8+6=14; right = 2×8 + 6×1 = 16+6=22? Wait — mistake.

Wait: H₃PO₄ has 4 O each → 2×4=8
Mg(OH)₂ has 2 O each → 3×2=6 → total O=14

Right: Mg₃(PO₄)₂ has 8 O (since PO₄ is 4 O ×2 = 8), H₂O has 6×1=6 → total 14 → OK.

✔ Balanced: 2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O

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10. NaOH + H₂CO₃ → Na₂CO₃ + H₂O

Carbonic acid + sodium hydroxide.

Na₂CO₃ needs 2 Na → so 2 NaOH.

Left: 2Na, 2O+2H from NaOH + 2H, 1C, 3O from H₂CO₃ → total H=4, O=5, Na=2, C=1

Right: Na₂CO₃ → 2Na, 1C, 3O; H₂O → 2H, 1O → total H=2, O=4 → mismatch.

Need 2 H₂O → then H=4, O=3+2=5 → perfect.

✔ Balanced: 2NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O

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11. KOH + HBr → KBr + H₂O

Acid-base reaction.

Already balanced!

Left: K=1, O=1, H=1+1=2, Br=1
Right: K=1, Br=1, H=2, O=1

✔ Balanced: KOH + HBr → KBr + H₂O

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12. H₂ + O₂ → H₂O₂

Hydrogen peroxide.

Left: H=2, O=2
Right: H=2, O=2

✔ Already balanced: H₂ + O₂ → H₂O₂

*(Note: This is not the typical combustion reaction — it’s direct synthesis of H₂O₂.)*

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13. Na + O₂ → Na₂O

Sodium oxide.

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

→ Put 4 Na on left → 2 Na₂O on right → O=2

✔ Balanced: 4Na + O₂ → 2Na₂O

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14. Al(OH)₃ + H₂CO₃ → Al₂(CO₃)₃ + H₂O

Aluminum hydroxide + carbonic acid → aluminum carbonate + water.

Al₂(CO₃)₃ requires 2 Al and 3 CO₃.

So need 2 Al(OH)₃ → gives 2 Al, 6 O, 6 H

Need 3 H₂CO₃ → gives 3 C, 9 O, 6 H

Total left: Al=2, C=3, O=6+9=15, H=6+6=12

Right: Al₂(CO₃)₃ → 2 Al, 3 C, 9 O
H₂O → need 6 H₂O → 12 H, 6 O → total O=9+6=15 → perfect.

✔ Balanced: 2Al(OH)₃ + 3H₂CO₃ → Al₂(CO₃)₃ + 6H₂O

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15. Al + S₈ → Al₂S₃

Aluminum sulfide.

S₈ has 8 S atoms.

Al₂S₃ has 3 S → need multiple of 3 and 8 → LCM is 24.

So 8 Al₂S₃ → 16 Al and 24 S

Requires 3 S₈ → 24 S

And 16 Al

✔ Balanced: 16Al + 3S₈ → 8Al₂S₃

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16. Cs + N₂ → Cs₃N

Cesium nitride.

Cs₃N has 3 Cs and 1 N.

N₂ has 2 N → so need 2 Cs₃N → 6 Cs and 2 N

Left: N₂ → 2 N → good

Need 6 Cs

✔ Balanced: 6Cs + N₂ → 2Cs₃N

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17. Mg + Cl₂ → MgCl₂

Already balanced!

Left: Mg=1, Cl=2
Right: Mg=1, Cl=2

✔ Balanced: Mg + Cl₂ → MgCl₂

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## 📋 Final Answer Summary:

Here are all 17 equations balanced:

1. 2H₂ + O₂ → 2H₂O
2. H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
3. 6K + B₂O₃ → 3K₂O + 2B
4. HCl + NaOH → NaCl + H₂O
5. 10Na + 2NaNO₃ → 6Na₂O + N₂
6. 4C + S₈ → 4CS₂
7. 2Na + O₂ → Na₂O₂
8. 2N₂ + 5O₂ → 2N₂O₅
9. 2H₃PO₄ + 3Mg(OH)₂ → Mg₃(PO₄)₂ + 6H₂O
10. 2NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O
11. KOH + HBr → KBr + H₂O
12. H₂ + O₂ → H₂O₂
13. 4Na + O₂ → 2Na₂O
14. 2Al(OH)₃ + 3H₂CO₃ → Al₂(CO₃)₃ + 6H₂O
15. 16Al + 3S₈ → 8Al₂S₃
16. 6Cs + N₂ → 2Cs₃N
17. Mg + Cl₂ → MgCl₂

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✔ All equations are now balanced according to the law of conservation of mass.

Let me know if you’d like to see the step-by-step atom counts for any specific equation!