Balancing chemical equations worksheet for Unit 3, Lesson 1, focusing on quantities in chemical reactions.

You’ve uploaded an image showing “Unit 2: Quantities in Chemical Reactions — Lesson 1: Balancing Chemical Equations”, which contains 10 unbalanced chemical equations. Your task is to balance each equation.

Balancing chemical equations means ensuring that the number of atoms of each element is the same on both sides (reactants and products), following the Law of Conservation of Mass.

Let’s solve each one step by step.

---

1. ___ NH₃ + ___ O₂ → ___ NO + ___ H₂O

Step-by-step:
- N: 1 on left, 1 on right → OK
- H: 3 on left, 2 on right → need to balance
- O: 2 on left, 1+1=2 on right? Wait — NO has 1 O, H₂O has 1 O → total 2 O on right, but let’s count properly after balancing.

Try coefficients:

Start with NH₃ → try 4 NH₃ to get even H’s.

4 NH₃ → 12 H → so need 6 H₂O (since each has 2 H)

Now:
4 NH₃ + ___ O₂ → ___ NO + 6 H₂O

N: 4 on left → need 4 NO
O: from 6 H₂O → 6 O; from 4 NO → 4 O → total 10 O on right → need 5 O₂ (since each O₂ gives 2 O)

✔ Final:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

---

2. ___ KClO₃ → ___ KCl + ___ O₂

Decomposition reaction.

K: 1 = 1
Cl: 1 = 1
O: 3 on left → need 3/2 O₂ → multiply all by 2 to eliminate fraction.

2 KClO₃ → 2 KCl + 3 O₂

✔ Balanced.

---

3. ___ NaOH + ___ Cl₂ → ___ NaCl + ___ NaClO + ___ H₂O

This is a disproportionation reaction of chlorine in base.

NaOH + Cl₂ → NaCl + NaClO + H₂O

Check atoms:

Left: Na=1, O=1, H=1, Cl=2
Right: Na=2, Cl=2, O=1, H=2 → not balanced.

Try 2 NaOH:

2 NaOH + Cl₂ → NaCl + NaClO + H₂O

Now:
Na: 2 = 2
O: 2 vs 1+1=2 → OK
H: 2 vs 2 → OK
Cl: 2 = 1+1 → OK

✔ Final:
2 NaOH + Cl₂ → NaCl + NaClO + H₂O

*(Note: This is the classic reaction for bleach formation.)*

---

4. ___ N₂ + ___ O₂ → ___ N₂O

N₂ + O₂ → N₂O

Left: N=2, O=2
Right: N=2, O=1 → need 2 N₂O → then N=4, O=2 → so need 2 N₂ on left.

2 N₂ + O₂ → 2 N₂O

✔ Balanced.

---

5. ___ Pb(NO₃)₂ + ___ HCl → ___ PbCl₂ + ___ HNO₃

Double displacement.

Pb(NO₃)₂ + HCl → PbCl₂ + HNO₃

Pb: 1 = 1
N: 2 on left → need 2 HNO₃
H: 2 on right → need 2 HCl
Cl: 2 on right → 2 HCl gives 2 Cl → OK

✔ Final:
Pb(NO₃)₂ + 2 HCl → PbCl₂ + 2 HNO₃

---

6. ___ AlBr₃ + ___ K₂SO₄ → ___ Al₂(SO₄)₃ + ___ KBr

Double displacement.

AlBr₃ + K₂SO₄ → Al₂(SO₄)₃ + KBr

Al: 1 vs 2 → need 2 AlBr₃
Br: 6 → need 6 KBr
K: 6 → need 3 K₂SO₄
SO₄: 3 → matches Al₂(SO₄)₃

✔ Final:
2 AlBr₃ + 3 K₂SO₄ → Al₂(SO₄)₃ + 6 KBr

---

7. ___ C₆H₆ + ___ O₂ → ___ CO₂ + ___ H₂O

Combustion of benzene.

C₆H₆ + O₂ → CO₂ + H₂O

C: 6 → need 6 CO₂
H: 6 → need 3 H₂O
O: from 6 CO₂ → 12 O; from 3 H₂O → 3 O → total 15 O → need 15/2 O₂ → multiply all by 2.

2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O

✔ Balanced.

---

8. ___ C₆H₁₄ + ___ O₂ → ___ CO₂ + ___ H₂O

Combustion of hexane.

C₆H₁₄ + O₂ → CO₂ + H₂O

C: 6 → 6 CO₂
H: 14 → 7 H₂O
O: 6×2 + 7×1 = 12 + 7 = 19 → need 19/2 O₂ → multiply by 2.

2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O

✔ Balanced.

---

9. ___ C₈H₁₈ + ___ O₂ → ___ CO₂ + ___ H₂O

Combustion of octane.

C₈H₁₈ + O₂ → CO₂ + H₂O

C: 8 → 8 CO₂
H: 18 → 9 H₂O
O: 8×2 + 9×1 = 16 + 9 = 25 → need 25/2 O₂ → multiply by 2.

2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

✔ Balanced.

---

10. ___ FeCl₃ + ___ KOH → ___ Fe(OH)₃ + ___ KCl

Double displacement.

FeCl₃ + KOH → Fe(OH)₃ + KCl

Fe: 1 = 1
Cl: 3 → need 3 KCl
K: 3 → need 3 KOH
O and H: 3 OH from 3 KOH → matches Fe(OH)₃

✔ Final:
FeCl₃ + 3 KOH → Fe(OH)₃ + 3 KCl

---

## ✔ FINAL ANSWERS (Balanced Equations):

1. 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
2. 2 KClO₃ → 2 KCl + 3 O₂
3. 2 NaOH + Cl₂ → NaCl + NaClO + H₂O
4. 2 N₂ + O₂ → 2 N₂O
5. Pb(NO₃)₂ + 2 HCl → PbCl₂ + 2 HNO₃
6. 2 AlBr₃ + 3 K₂SO₄ → Al₂(SO₄)₃ + 6 KBr
7. 2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O
8. 2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O
9. 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
10. FeCl₃ + 3 KOH → Fe(OH)₃ + 3 KCl

---

## 💡 Explanation Summary:

- Always start by counting atoms of each element on both sides.
- Use coefficients (whole numbers) to balance — never change subscripts.
- For combustion reactions (hydrocarbons + O₂ → CO₂ + H₂O), balance C first, then H, then O.
- For double displacement or decomposition, balance ions or polyatomic groups if possible.
- If you get fractions, multiply the entire equation by the denominator to clear them.

Let me know if you want to see the atom counts for any specific equation!