Balancing Equations Practice: Three combustion reactions to balance, featuring CH₄, C₃H₈, and C₈H₁₈ with O₂, CO₂, and H₂O.

Let’s balance each equation one at a time. We’ll count atoms on both sides and adjust coefficients (the numbers in front) until everything matches.

---

First Equation: CH₄ + O₂ → CO₂ + H₂O

Left side:
- C: 1 (from CH₄)
- H: 4 (from CH₄)
- O: 2 (from O₂)

Right side:
- C: 1 (from CO₂) ✔
- H: 2 (from H₂O) ✘ need 4
- O: 3 (2 from CO₂ + 1 from H₂O) ✘

Step 1: Fix hydrogen. Put a 2 in front of H₂O → now H = 4 on right.

Now: CH₄ + O₂ → CO₂ + 2H₂O

Right side O: 2 (from CO₂) + 2×1 = 4 total oxygen atoms.

Left side O: still only 2 (from O₂). So we need 4 oxygen atoms on left → put a 2 in front of O₂.

Now: CH₄ + 2O₂ → CO₂ + 2H₂O

Check all:

Left: C=1, H=4, O=4
Right: C=1, H=4, O=2+2=4 ✔

✔ Balanced! Coefficients: 1, 2, 1, 2

---

Second Equation: C₃H₈ + O₂ → CO₂ + H₂O

Left:
- C: 3
- H: 8
- O: 2

Right:
- C: 1 → need 3 → put 3 in front of CO₂
- H: 2 → need 8 → put 4 in front of H₂O (since 4×2=8)

Try: C₃H₈ + O₂ → 3CO₂ + 4H₂O

Now check oxygen on right:
3×2 = 6 from CO₂
4×1 = 4 from H₂O
Total O = 10

So left needs 10 oxygen atoms → O₂ has 2 per molecule → need 5 O₂ molecules.

Final: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Check:

Left: C=3, H=8, O=10
Right: C=3, H=8, O=6+4=10 ✔

✔ Balanced! Coefficients: 1, 5, 3, 4

---

Third Equation: C₈H₁₈ + O₂ → CO₂ + H₂O

Left:
- C: 8
- H: 18
- O: 2

Right:
- C: 1 → need 8 → put 8 in front of CO₂
- H: 2 → need 18 → put 9 in front of H₂O (9×2=18)

Try: C₈H₁₈ + O₂ → 8CO₂ + 9H₂O

Oxygen on right:
8×2 = 16 from CO₂
9×1 = 9 from H₂O
Total O = 25

Left: O₂ gives 2 per molecule → need 25/2 = 12.5 → not whole number!

We can’t have half molecules. Multiply entire equation by 2 to eliminate fraction.

Start over with doubled hydrocarbon:

Use 2C₈H₁₈ → then C = 16, H = 36

Then:
→ 16 CO₂ (for carbon)
→ 18 H₂O (for hydrogen: 18×2=36)

Now oxygen on right:
16×2 = 32 from CO₂
18×1 = 18 from H₂O
Total O = 50

So left needs 50 oxygen atoms → O₂ → 50 ÷ 2 = 25

Equation: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Check:

Left: C=16, H=36, O=50
Right: C=16, H=36, O=32+18=50 ✔

But usually we want smallest whole numbers. Can we divide all by 2?

2, 25, 16, 18 → 25 is odd → can’t divide evenly.

So this is the simplest balanced form.

Wait — actually, let’s double-check if there’s a smaller set.

Original attempt gave us fractional coefficient for O₂: 12.5 → so multiplying by 2 was correct.

Yes, 2, 25, 16, 18 is the lowest whole-number ratio.

✔ Balanced! Coefficients: 2, 25, 16, 18

---

Final Answer:
For CH₄ + O₂ → CO₂ + H₂O: 1, 2, 1, 2
For C₃H₈ + O₂ → CO₂ + H₂O: 1, 5, 3, 4
For C₈H₁₈ + O₂ → CO₂ + H₂O: 2, 25, 16, 18