Solved Balancing Chemical Equations - Lab 7 Balancing | Chegg.com - Free Printable

Let's go through each of these chemical equations and balance them step by step. Balancing chemical equations means making sure the number of atoms of each element is the same on both sides of the equation (conservation of mass).

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1) Br₂ + LiF → LiBr + F₂

Unbalanced:
Br₂ + LiF → LiBr + F₂

We need to balance:
- Br: 2 on left, 1 on right
- F: 1 on left, 2 on right
- Li: 1 on both sides

Start with Br: put 2 LiBr on right → now 2 Br on right
Then we need 2 LiF on left → gives 2 Li and 2 F
But F₂ has 2 F, so we need 1 F₂ on right

Now check:
Left: Br₂ (2 Br), 2 LiF (2 Li, 2 F)
Right: 2 LiBr (2 Li, 2 Br), F₂ (2 F)

✔ Balanced.

Answer:
1 Br₂ + 2 LiF → 2 LiBr + 1 F₂

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2) H₃PO₄ + Fe(OH)₂ → H₂O + Fe₃(PO₄)₂

Unbalanced:
H₃PO₄ + Fe(OH)₂ → H₂O + Fe₃(PO₄)₂

Look at Fe: 3 on right, 1 on left → need 3 Fe(OH)₂
PO₄: 2 on right, 1 on left → need 2 H₃PO₄

So far:
2 H₃PO₄ + 3 Fe(OH)₂ → H₂O + Fe₃(PO₄)₂

Now count H and O:

Left:
- H: from 2 H₃PO₄ → 6 H, from 3 Fe(OH)₂ → 3×2 = 6 H → total 12 H
- O: 2×4 = 8 from H₃PO₄, 3×2 = 6 from Fe(OH)₂ → total 14 O

Right:
Fe₃(PO₄)₂ has no H or O in its own structure, but H₂O has H and O

So we need to produce water. Each H₂O has 2 H and 1 O.

Total H: 12 → need 6 H₂O
Total O: 6 H₂O → 6 O, plus Fe₃(PO₄)₂ has 8 O → total 14 O ✔

Check:
Left: 2 H₃PO₄ + 3 Fe(OH)₂
Right: Fe₃(PO₄)₂ + 6 H₂O

Atoms:
- P: 2 = 2 ✔
- Fe: 3 = 3 ✔
- H: 6 (from H₃PO₄) + 6 (from Fe(OH)₂) = 12 → 6 H₂O → 12 H ✔
- O: 8 (from H₃PO₄) + 6 (from Fe(OH)₂) = 14 → 6 H₂O → 6 O, Fe₃(PO₄)₂ → 8 O → total 14 ✔

Answer:
2 H₃PO₄ + 3 Fe(OH)₂ → 6 H₂O + 1 Fe₃(PO₄)₂

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3) C₃H₈OH + O₂ → CO₂ + H₂O

This is combustion of propanol (C₃H₈OH). General form: hydrocarbon + O₂ → CO₂ + H₂O

C₃H₈OH has 3 C, 8 H, 1 O (plus OH group)

So:
- C: 3 → 3 CO₂
- H: 8 → 4 H₂O (since each has 2 H)
- O: Left: 1 from fuel + 2 per O₂ molecule; Right: 3×2 = 6 from CO₂ + 4×1 = 4 from H₂O → total 10 O

So O needed: 10 - 1 = 9 → 9/2 = 4.5 O₂ → multiply whole equation by 2

Start over with coefficients:

Try:
C₃H₈OH + 9/2 O₂ → 3 CO₂ + 4 H₂O

Multiply all by 2:
2 C₃H₈OH + 9 O₂ → 6 CO₂ + 8 H₂O

Check:
- C: 6 = 6 ✔
- H: 2×8 = 16 → 8 H₂O → 16 H ✔
- O: Left: 2×1 (from fuel) + 9×2 = 18 → total 20 O
Right: 6×2 = 12 (CO₂) + 8×1 = 8 (H₂O) → 20 ✔

Answer:
2 C₃H₈OH + 9 O₂ → 6 CO₂ + 8 H₂O

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4) Ni(OH)₃ → Ni₂O₃ + H₂O

Decomposition.

Left: Ni(OH)₃ → 1 Ni, 3 O, 3 H
Right: Ni₂O₃ → 2 Ni, 3 O; H₂O → 2 H, 1 O

Need 2 Ni on right → use 2 Ni(OH)₃

2 Ni(OH)₃ → Ni₂O₃ + H₂O

Now H: 2×3 = 6 H → need 3 H₂O

O: Left: 2×3 = 6 O
Right: Ni₂O₃ → 3 O, 3 H₂O → 3 O → total 6 ✔

H: 6 H → 3 H₂O → 6 H ✔

Answer:
2 Ni(OH)₃ → 1 Ni₂O₃ + 3 H₂O

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5) K₂SO₃ + Mn(OH)₂ → KOH + MnSO₃

Double displacement.

Left: K₂SO₃ → 2 K, S, 3 O
Mn(OH)₂ → Mn, 2 O, 2 H

Right: KOH → K, O, H
MnSO₃ → Mn, S, 3 O

So:
- K: 2 on left → need 2 KOH
- Mn: 1 → 1 MnSO₃
- SO₃: 1 → 1 MnSO₃
- H: 2 on left → 2 KOH → 2 H → matches

Check O:
Left: 3 (K₂SO₃) + 2 (Mn(OH)₂) = 5 O
Right: 2 KOH → 2 O, MnSO₃ → 3 O → 5 O ✔

H: 2 → 2 ✔

Answer:
1 K₂SO₃ + 1 Mn(OH)₂ → 2 KOH + 1 MnSO₃

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6) NaOH + H₂SO₄ → H₂O + Na₂SO₄

Acid-base neutralization.

NaOH + H₂SO₄ → H₂O + Na₂SO₄

Na: 1 left, 2 right → need 2 NaOH
H: 2 from H₂SO₄ + 2 from 2 NaOH = 4 H → 2 H₂O

Check:
2 NaOH + H₂SO₄ → 2 H₂O + Na₂SO₄

Atoms:
- Na: 2 = 2 ✔
- O: 2 (NaOH) + 4 (H₂SO₄) = 6 → 2 H₂O → 2 O, Na₂SO₄ → 4 O → 6 ✔
- H: 2 + 2 = 4 → 2 H₂O → 4 H ✔
- S: 1 = 1 ✔

Answer:
2 NaOH + 1 H₂SO₄ → 2 H₂O + 1 Na₂SO₄

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7) Li + Pb(OH)₂ → Pb + LiOH

Single displacement.

Pb(OH)₂ → Pb²⁺ and 2 OH⁻
Li → Li⁺

So need 2 Li to balance charge.

2 Li + Pb(OH)₂ → Pb + 2 LiOH

Check:
- Li: 2 = 2 ✔
- Pb: 1 = 1 ✔
- O: 2 = 2 ✔
- H: 2 = 2 ✔

Answer:
2 Li + 1 Pb(OH)₂ → 1 Pb + 2 LiOH

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8) C₄H₈ + O₂ → CO₂ + H₂O

Combustion of butene.

C₄H₈ → 4 C, 8 H

→ 4 CO₂, 4 H₂O (since 8 H → 4 H₂O)

O needed: 4 CO₂ → 8 O, 4 H₂O → 4 O → total 12 O → 6 O₂

So:
C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

Check:
- C: 4 = 4 ✔
- H: 8 = 8 ✔
- O: 12 = 8 + 4 = 12 ✔

Answer:
1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

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9) Ga(OH)₃ + KF → KOH + GaF₃

Double displacement.

Ga(OH)₃ → Ga³⁺, 3 OH⁻
KF → K⁺, F⁻

To get GaF₃, need 3 F⁻ → 3 KF
Then 3 KOH

So:
Ga(OH)₃ + 3 KF → 3 KOH + GaF₃

Check:
- Ga: 1 = 1 ✔
- O: 3 = 3 ✔
- H: 3 = 3 ✔
- K: 3 = 3 ✔
- F: 3 = 3 ✔

Answer:
1 Ga(OH)₃ + 3 KF → 3 KOH + 1 GaF₃

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10) V + ZnBr₂ → VBr₃ + Zn

Single displacement.

V → V³⁺ → needs 3 Br⁻ → VBr₃
ZnBr₂ → Zn²⁺, 2 Br⁻

So need to balance Br.

VBr₃ has 3 Br, ZnBr₂ has 2 Br → LCM = 6

So: 2 VBr₃ → 6 Br → need 3 ZnBr₂

Then Zn: 3 Zn produced → need 3 Zn on left? No — Zn is product.

Reactants: V and ZnBr₂

3 ZnBr₂ → 3 Zn + 6 Br⁻ → 2 VBr₃ → 2 V

So:
2 V + 3 ZnBr₂ → 2 VBr₃ + 3 Zn

Check:
- V: 2 = 2 ✔
- Zn: 3 = 3 ✔
- Br: 6 = 6 ✔

Answer:
2 V + 3 ZnBr₂ → 2 VBr₃ + 3 Zn

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11) As₂O₅ + H₂O → H₃AsO₄

Hydrolysis.

As₂O₅ → 2 As, 5 O
H₃AsO₄ → 1 As, 4 O, 3 H

So need 2 H₃AsO₄ → 2 As, 8 O, 6 H

Left: As₂O₅ → 2 As, 5 O
Need 3 more O and 6 H → H₂O provides 2 H and 1 O per molecule

Need 3 H₂O → 3 O and 6 H

Total O: 5 + 3 = 8 → matches 2 H₃AsO₄ → 8 O

So:
As₂O₅ + 3 H₂O → 2 H₃AsO₄

Check:
- As: 2 = 2 ✔
- O: 5 + 3 = 8 → 2×4 = 8 ✔
- H: 6 = 6 ✔

Answer:
1 As₂O₅ + 3 H₂O → 2 H₃AsO₄

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12) NH₃ + O₂ → NO + H₂O

Combustion of ammonia.

NH₃ → N, 3 H
NO → N, O
H₂O → 2 H, 1 O

Balance N: 1 each side → OK

H: 3 on left → need 3/2 H₂O → 1.5 H₂O

O: Right: 1 (NO) + 1.5 (H₂O) = 2.5 O → need 2.5/2 = 1.25 O₂

Multiply by 4 to eliminate fractions:

Original:
NH₃ + 1.25 O₂ → NO + 1.5 H₂O

×4:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Check:
- N: 4 = 4 ✔
- H: 12 = 12 ✔
- O: 10 = 4 (NO) + 6 (H₂O) = 10 ✔

Answer:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

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13) C₃H₆ + O₂ → CO₂ + H₂O

Combustion of propene.

C₃H₆ → 3 C, 6 H

→ 3 CO₂, 3 H₂O

O needed: 3×2 = 6 (CO₂) + 3×1 = 3 (H₂O) = 9 O → 4.5 O₂ → ×2

So:
2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O

Check:
- C: 6 = 6 ✔
- H: 12 = 12 ✔
- O: 18 = 12 + 6 = 18 ✔

Answer:
2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O

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14) NaClO₃ → NaCl + O₂

Decomposition.

NaClO₃ → Na, Cl, 3 O
NaCl → Na, Cl
O₂ → 2 O

So O: 3 → need 3/2 O₂ → ×2

2 NaClO₃ → 2 NaCl + 3 O₂

Check:
- Na: 2 = 2 ✔
- Cl: 2 = 2 ✔
- O: 6 = 6 ✔

Answer:
2 NaClO₃ → 2 NaCl + 3 O₂

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15) Ca + O₂ → CaO

Calcium oxide formation.

Ca + O₂ → CaO

O: 2 on left, 1 on right → need 2 CaO
Then Ca: 2 on right → need 2 Ca

So:
2 Ca + O₂ → 2 CaO

Check:
- Ca: 2 = 2 ✔
- O: 2 = 2 ✔

Answer:
2 Ca + 1 O₂ → 2 CaO

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## ✔ Final Answers:

1) 1 Br₂ + 2 LiF → 2 LiBr + 1 F₂
2) 2 H₃PO₄ + 3 Fe(OH)₂ → 6 H₂O + 1 Fe₃(PO₄)₂
3) 2 C₃H₈OH + 9 O₂ → 6 CO₂ + 8 H₂O
4) 2 Ni(OH)₃ → 1 Ni₂O₃ + 3 H₂O
5) 1 K₂SO₃ + 1 Mn(OH)₂ → 2 KOH + 1 MnSO₃
6) 2 NaOH + 1 H₂SO₄ → 2 H₂O + 1 Na₂SO₄
7) 2 Li + 1 Pb(OH)₂ → 1 Pb + 2 LiOH
8) 1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O
9) 1 Ga(OH)₃ + 3 KF → 3 KOH + 1 GaF₃
10) 2 V + 3 ZnBr₂ → 2 VBr₃ + 3 Zn
11) 1 As₂O₅ + 3 H₂O → 2 H₃AsO₄
12) 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
13) 2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O
14) 2 NaClO₃ → 2 NaCl + 3 O₂
15) 2 Ca + 1 O₂ → 2 CaO

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