Honors Chemistry worksheet on writing and balancing chemical equations, featuring problems and equations to solve.

Let’s go step by step to solve each problem. We’ll write the chemical equations first, then balance them.

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Part 1: Write and Balance Chemical Equations from Word Descriptions

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Problem 1:
*When lithium hydroxide pellets are added to a solution of sulfuric acid (H₂SO₄), lithium sulfate and water are formed.*

- Reactants: LiOH + H₂SO₄
- Products: Li₂SO₄ + H₂O

Unbalanced:
LiOH + H₂SO₄ → Li₂SO₄ + H₂O

Balance Li: Need 2 Li on left → 2 LiOH
Now H: Left = 2 (from 2 LiOH) + 2 (from H₂SO₄) = 4 H; Right = 2 H in H₂O → need 2 H₂O
Check O and S — already balanced.

✔ Balanced:
2 LiOH + H₂SO₄ → Li₂SO₄ + 2 H₂O

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Problem 2:
*Magnesium reacts with sodium fluoride to produce magnesium fluoride and elemental sodium.*

- Reactants: Mg + NaF
- Products: MgF₂ + Na

Unbalanced:
Mg + NaF → MgF₂ + Na

Balance F: Need 2 F on left → 2 NaF
Then Na: 2 Na on left → need 2 Na on right

✔ Balanced:
Mg + 2 NaF → MgF₂ + 2 Na

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Problem 3:
*If a copper coil is placed into a solution of silver nitrate, silver crystals form and copper (I) nitrate is generated.*

Note: Copper (I) means Cu⁺, so copper(I) nitrate is CuNO₃
Silver nitrate is AgNO₃
Silver metal is Ag
Copper metal is Cu

Reactants: Cu + AgNO₃
Products: Ag + CuNO₃

Unbalanced:
Cu + AgNO₃ → Ag + CuNO₃

Check atoms:
Left: Cu=1, Ag=1, N=1, O=3
Right: Ag=1, Cu=1, N=1, O=3 → Already balanced!

✔ Balanced:
Cu + AgNO₃ → Ag + CuNO₃

*(Note: In reality, copper usually forms Cu²⁺, but the problem says “copper (I)”, so we follow that.)*

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Problem 4:
*When crystalline C₆H₁₂O₆ is burned in oxygen, carbon dioxide and water vapor are formed.*

This is combustion of glucose.

Reactants: C₆H₁₂O₆ + O₂
Products: CO₂ + H₂O

Unbalanced:
C₆H₁₂O₆ + O₂ → CO₂ + H₂O

Balance C: 6 CO₂
Balance H: 12 H → 6 H₂O
Now count O on right: 6×2 + 6×1 = 12 + 6 = 18 O
Left: C₆H₁₂O₆ has 6 O → need 12 more O from O₂ → 6 O₂ molecules

✔ Balanced:
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

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Problem 5:
*Calcium carbonate combines with hydrochloric acid (HCl) to produce calcium chloride, water and carbon dioxide gas.*

Reactants: CaCO₃ + HCl
Products: CaCl₂ + H₂O + CO₂

Unbalanced:
CaCO₃ + HCl → CaCl₂ + H₂O + CO₂

Balance Cl: Need 2 HCl for CaCl₂
Then H: 2 H → makes 1 H₂O
Check all: Ca, C, O also balanced.

✔ Balanced:
CaCO₃ + 2 HCl → CaCl₂ + H₂O + CO₂

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Part 2: Balance the Given Equations

We’ll fill in coefficients to make atom counts equal on both sides.

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1) ___ N₂ + ___ H₂ → ___ NH₃

N: 2 on left → need 2 NH₃ → 2 N, 6 H
H₂: need 3 H₂ to get 6 H

✔ 1 N₂ + 3 H₂ → 2 NH₃

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2) ___ KClO₃ → ___ KCl + ___ O₂

K and Cl: 1 each side if 1:1
O: 3 on left → need 3/2 O₂ on right → multiply all by 2

✔ 2 KClO₃ → 2 KCl + 3 O₂

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3) ___ NaCl + ___ F₂ → ___ NaF + ___ Cl₂

Na: 1 each → ok
Cl: 1 on left, 2 on right → need 2 NaCl
Then Na: 2 → need 2 NaF
F: 2 on left (F₂), 2 on right (2 NaF) → good
Cl: 2 on left, 2 on right (Cl₂)

✔ 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂

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4) ___ H₂ + ___ O₂ → ___ H₂O

H: 2 on left, 2 on right → ok
O: 2 on left, 1 on right → need 2 H₂O → then H becomes 4 → need 2 H₂

✔ 2 H₂ + 1 O₂ → 2 H₂O

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5) ___ Pb(OH)₂ + ___ HCl → ___ H₂O + ___ PbCl₂

Pb: 1 each
Cl: 1 on left, 2 on right → need 2 HCl
H: Left: 2 (from Pb(OH)₂) + 2 (from 2 HCl) = 4 H → need 2 H₂O
O: 2 on left → 2 on right (in 2 H₂O)

✔ 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂

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

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

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

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7) ___ CH₄ + ___ O₂ → ___ CO₂ + ___ H₂O

C: 1 → 1
H: 4 → need 2 H₂O
O: right = 2 + 2 = 4 → need 2 O₂

✔ 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

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8) ___ C₃H₈ + ___ O₂ → ___ CO₂ + ___ H₂O

C: 3 → 3 CO₂
H: 8 → 4 H₂O
O: right = 6 + 4 = 10 → need 5 O₂

✔ 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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9) ___ C₈H₁ + ___ O₂ → ___ CO₂ + ___ H₂O

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

→ 2 C₈H₁ + 25 O₂ → 16 CO₂ + 18 H₂O

✔ 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

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10) ___ FeCl₃ + ___ NaOH → ___ Fe(OH)₃ + ___ NaCl

Fe: 1 → 1
Cl: 3 → need 3 NaCl
Na: 3 → need 3 NaOH
OH: 3 → matches Fe(OH)₃

✔ 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl

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11) ___ P + ___ O₂ → ___ P₂O₅

P: 1 vs 2 → need 2 P
O: 5 → need 5/2 O₂ → multiply all by 2

→ 4 P + 5 O₂ → 2 P₂O₅

✔ 4 P + 5 O₂ → 2 P₂O₅

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Final Answer:

Part 1: Word Problems

1) 2 LiOH + H₂SO₄ → Li₂SO₄ + 2 H₂O
2) Mg + 2 NaF → MgF₂ + 2 Na
3) Cu + AgNO₃ → Ag + CuNO₃
4) C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
5) CaCO₃ + 2 HCl → CaCl₂ + H₂O + CO₂

Part 2: Balancing Equations

1) 1 N₂ + 3 H₂ → 2 NH₃
2) 2 KClO₃ → 2 KCl + 3 O₂
3) 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂
4) 2 H₂ + 1 O₂ → 2 H₂O
5) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
6) 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
7) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
8) 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
9) 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
10) 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl
11) 4 P + 5 O₂ → 2 P₂O₅