Balancing Equations Worksheet with answers for chemistry practice.

It looks like you've uploaded a Balancing Chemical Equations Worksheet with 20 problems, and the answers for the first three are already provided. However, your request is to solve the problem and explain the solution, so I’ll go through each of the first few equations in detail, showing how to balance them step-by-step. Then I'll provide the full balanced answers for all 20 equations.

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🔬 How to Balance Chemical Equations: Step-by-Step Method

To balance a chemical equation:

1. Write the unbalanced equation.
2. Count atoms of each element on both sides.
3. Use coefficients (numbers in front of compounds) to balance atoms.
4. Start with elements that appear in only one compound on each side.
5. Balance polyatomic ions as groups if they remain unchanged.
6. Check all atoms again.
7. Simplify coefficients if possible (reduce to lowest whole numbers).

Let’s go through each problem.

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## ✔ Problem 1:
Na₃PO₄ + KOH → NaOH + K₃PO₄

Step-by-step:

Unbalanced:
Na₃PO₄ + KOH → NaOH + K₃PO₄

Count atoms:

| Element | Left Side | Right Side |
|--------|----------|-----------|
| Na | 3 | 1 |
| P | 1 | 1 |
| O | 4 + 1 = 5 | 1 + 4 = 5 |
| K | 1 | 3 |
| H | 1 | 1 |

We see:
- Na: 3 on left, 1 on right → need 3 NaOH
- K: 1 on left, 3 on right → need 3 KOH
- So try:
Na₃PO₄ + 3KOH → 3NaOH + K₃PO₄

Now check:

| Element | Left | Right |
|--------|------|-------|
| Na | 3 | 3 |
| P | 1 | 1 |
| O | 4 + 3 = 7 | 3 + 4 = 7 |
| K | 3 | 3 |
| H | 3 | 3 |

✔ Balanced!

Answer:
1 Na₃PO₄ + 3 KOH → 3 NaOH + 1 K₃PO₄

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## ✔ Problem 2:
MgF₂ + Li₂CO₃ → MgCO₃ + LiF

Unbalanced:
MgF₂ + Li₂CO₃ → MgCO₃ + LiF

Count atoms:

| Element | Left | Right |
|--------|------|-------|
| Mg | 1 | 1 |
| F | 2 | 1 |
| Li | 2 | 1 |
| C | 1 | 1 |
| O | 3 | 3 |

Li and F are unbalanced.

- F: 2 on left, 1 on right → need 2 LiF
- Li: 2 on left → matches 2 LiF
- So:
MgF₂ + Li₂CO₃ → MgCO₃ + 2LiF

Now check:

| Element | Left | Right |
|--------|------|-------|
| Mg | 1 | 1 |
| F | 2 | 2 |
| Li | 2 | 2 |
| C | 1 | 1 |
| O | 3 | 3 |

✔ Balanced!

Answer:
1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF

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## ✔ Problem 3:
P₄ + O₂ → P₂O₅

Unbalanced:
P₄ + O₂ → P₂O₅

Note: P₂O₅ is actually P₄O₁₀ in its standard form, but we'll follow the given formula.

But let's use P₂O₅ as written.

So:
P₄ + O₂ → P₂O₅

Atoms:

| Element | Left | Right |
|--------|------|-------|
| P | 4 | 2 |
| O | 2 | 5 |

Need to balance P and O.

- P: 4 on left → need 2 P₂O₅ (since each has 2 P)
- So: P₄ + O₂ → 2P₂O₅

Now O: Right = 2 × 5 = 10 → need 5 O₂ (each gives 2 O)

So:
P₄ + 5O₂ → 2P₂O₅

Check:

| Element | Left | Right |
|--------|------|-------|
| P | 4 | 4 |
| O | 10 | 10 |

✔ Balanced!

Answer:
1 P₄ + 5 O₂ → 2 P₂O₅

Wait — the answer sheet says:
1 P₄ + 3 O₂ → 2 P₂O₃

That suggests the product is P₂O₃, not P₂O₅.

Ah! There's a discrepancy.

Let’s check:
If product is P₂O₃, then:

P₄ + O₂ → P₂O₃

P: 4 → need 2 P₂O₃
O: 2×3 = 6 → need 3 O₂

So:
P₄ + 3 O₂ → 2 P₂O₃

This matches the answer key.

But note: P₂O₃ is phosphorus trioxide, while P₂O₅ is pentoxide. The correct formula depends on the reaction.

Since the worksheet says P₂O₃, we’ll go with it.

✔ Final Answer:
1 P₄ + 3 O₂ → 2 P₂O₃

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## 🧩 Let’s solve the rest quickly using same logic.

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4) RbNO₃ + BeF₂ → Be(NO₃)₂ + RbF

- Be: 1 on both sides → ok
- NO₃: 1 on left, 2 on right → need 2 RbNO₃
- Rb: 2 on left → need 2 RbF
- F: 2 on left → 2 on right → good

So:
2 RbNO₃ + 1 BeF₂ → 1 Be(NO₃)₂ + 2 RbF

✔ Balanced.

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5) AgNO₃ + Cu → Cu(NO₃)₂ + Ag

- Cu: 1 on left, 1 on right → ok
- NO₃: 1 on left, 2 on right → need 2 AgNO₃
- Ag: 2 on left → need 2 Ag on right

So:
2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag

✔ Balanced.

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6) CF₄ + Br₂ → CBr₄ + F₂

- C: 1 on both sides → ok
- F: 4 on left → need 2 F₂ on right
- Br: 2 on left → need 1 CBr₄ → 4 Br needed → so need 2 Br₂

So:
CF₄ + 2Br₂ → CBr₄ + 2F₂

Check:
F: 4 = 4
Br: 4 = 4
C: 1 = 1

✔ Balanced.

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7) HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂

- Cu: 1 → 1 → ok
- SO₄: 1 → 1 → ok
- H: 1 on left, 2 on right → need 2 HCN
- CN: 2 on left → need 1 Cu(CN)₂ → CN: 2 → ok

So:
2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂

✔ Balanced.

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8) GaF₃ + Cs → CsF + Ga

- Ga: 1 → 1 → ok
- F: 3 → 1 → need 3 CsF → so 3 Cs
- Cs: 3 on right → need 3 Cs on left

So:
1 GaF₃ + 3 Cs → 3 CsF + 1 Ga

✔ Balanced.

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9) BaS + PtF₂ → BaF₂ + PtS

- Ba: 1 → 1 → ok
- S: 1 → 1 → ok
- Pt: 1 → 1 → ok
- F: 2 → 2 → ok

All atoms balanced as is.

✔ 1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS

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10) N₂ + H₂ → NH₃

Classic ammonia synthesis.

- N₂ → 2 N → need 2 NH₃
- H: 6 on right → need 3 H₂

So:
1 N₂ + 3 H₂ → 2 NH₃

✔ Balanced.

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11) NaF + Br₂ → NaBr + F₂

- Na: 1 → 1 → ok
- F: 1 → 2 → need 2 NaF
- Br: 2 → 2 → need 2 NaBr

So:
2 NaF + 1 Br₂ → 2 NaBr + 1 F₂

✔ Balanced.

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12) Pb(OH)₂ + HCl → H₂O + PbCl₂

- Pb: 1 → 1 → ok
- OH: 2 → need 2 H₂O
- H: from OH and HCl → total H on left: 2 (from OH) + 1 (from HCl) → wait

Better:
Pb(OH)₂ has 2 OH groups → can react with 2 H⁺

So:
Pb(OH)₂ + 2 HCl → PbCl₂ + 2 H₂O

Check:
- Pb: 1 → 1
- O: 2 → 2 (in H₂O)
- H: 2 (from OH) + 2 (from HCl) = 4 → 2 H₂O has 4 H → ok
- Cl: 2 → 2

✔ Balanced.

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13) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

- Al: 1 → 2 → need 2 AlBr₃
- SO₄: 1 → 3 → need 3 K₂SO₄
- K: 6 → need 6 KBr
- Br: 6 → 6

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

✔ Balanced.

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14) CH₄ + O₂ → CO₂ + H₂O

Combustion of methane.

- C: 1 → 1 → ok
- H: 4 → 2 in H₂O → need 2 H₂O
- O: 2 (from O₂) → CO₂ has 2, H₂O has 1 → total 3 → need 2 O₂ (4 O)

So:
CH₄ + 2O₂ → CO₂ + 2H₂O

Check:
- C: 1 → 1
- H: 4 → 4
- O: 4 → 2 + 2 = 4

✔ Balanced.

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15) Na₃PO₄ + CaCl₂ → NaCl + Ca₃(PO₄)₂

- PO₄: 1 → 2 → need 2 Na₃PO₄
- Ca: 1 → 3 → need 3 CaCl₂
- Na: 6 → need 6 NaCl
- Cl: 6 → 6

So:
2 Na₃PO₄ + 3 CaCl₂ → 6 NaCl + 1 Ca₃(PO₄)₂

✔ Balanced.

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16) K + Cl₂ → KCl

- K: 1 → 1 → ok
- Cl: 2 → 1 → need 2 KCl → need 2 K

So:
2 K + 1 Cl₂ → 2 KCl

✔ Balanced.

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17) Al + HCl → H₂ + AlCl₃

- Al: 1 → 1 → ok
- Cl: 1 → 3 → need 3 HCl
- H: 3 → 2 → need 3/2 H₂ → multiply by 2

So:
2 Al + 6 HCl → 3 H₂ + 2 AlCl₃

✔ Balanced.

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18) N₂ + F₂ → NF₃

- N: 2 → 1 → need 2 NF₃
- F: 6 → need 3 F₂

So:
1 N₂ + 3 F₂ → 2 NF₃

✔ Balanced.

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19) SO₂ + Li₂Se → SSE₂ + Li₂O

- S: 1 → 1 → ok
- O: 2 → 1 → need 2 Li₂O
- Li: 2 → 4 → need 2 Li₂Se
- Se: 2 → 2 → ok

So:
1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O

Wait: SSe₂ has 1 S and 2 Se → yes

Check:
- S: 1 → 1
- O: 2 → 2
- Li: 4 → 4
- Se: 2 → 2

✔ Balanced.

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20) NH₃ + H₂SO₄ → (NH₄)₂SO₄

- N: 1 → 2 → need 2 NH₃
- H: 6 (from 2 NH₃) + 2 (from H₂SO₄) = 8 → (NH₄)₂SO₄ has 8 H → ok
- S: 1 → 1
- O: 4 → 4

So:
2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄

✔ Balanced.

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## ✔ Final Answers (Balanced):

1) 1 Na₃PO₄ + 3 KOH → 3 NaOH + 1 K₃PO₄
2) 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
3) 1 P₄ + 3 O₂ → 2 P₂O₃
4) 2 RbNO₃ + 1 BeF₂ → 1 Be(NO₃)₂ + 2 RbF
5) 2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag
6) 1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂
7) 2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂
8) 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
9) 1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS
10) 1 N₂ + 3 H₂ → 2 NH₃
11) 2 NaF + 1 Br₂ → 2 NaBr + 1 F₂
12) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
13) 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
14) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
15) 2 Na₃PO₄ + 3 CaCl₂ → 6 NaCl + 1 Ca₃(PO₄)₂
16) 2 K + 1 Cl₂ → 2 KCl
17) 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃
18) 1 N₂ + 3 F₂ → 2 NF₃
19) 1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O
20) 2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄

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📝 Summary:

The key to balancing equations is conservation of mass: atoms must be equal on both sides. Start with the most complex molecule, balance metals first, then nonmetals, and finally hydrogen and oxygen. Use coefficients (not subscripts!) to balance.

Let me know if you'd like this as a downloadable PDF or want explanations for specific ones!