Worksheet for balancing the combustion of gaseous butane (C₄H₁₀) with oxygen to form carbon dioxide and water.

Let's solve the problem step by step based on the worksheet you provided.

---

Problem:

Write a balanced equation for the combustion of gaseous butane (C₄H₁₀), which combines with gaseous oxygen (O₂) to form gaseous carbon dioxide (CO₂) and gaseous water (H₂O).

---

Step 1: Write the unbalanced chemical equation

We start by writing the correct formulas for all reactants and products:

- Butane: C₄H₁₀ (g)
- Oxygen: O₂ (g)
- Carbon dioxide: CO₂ (g)
- Water: H₂O (g)

So, the unbalanced equation is:

> C₄H₁₀ (g) + O₂ (g) → CO₂ (g) + H₂O (g)

---

Step 2: Balance the equation using the method described in the worksheet

We follow the steps:

#### Begin with C (Carbon)

There are 4 carbon atoms in C₄H₁₀, so we need 4 CO₂ molecules on the product side:

> C₄H₁₀ + O₂ → 4 CO₂ + H₂O

#### Balance H (Hydrogen)

There are 10 hydrogen atoms in C₄H₁₀. Each water molecule (H₂O) has 2 hydrogen atoms, so we need 5 H₂O molecules to get 10 hydrogens:

> C₄H₁₀ + O₂ → 4 CO₂ + 5 H₂O

#### Balance O (Oxygen)

Now count oxygen atoms on both sides:

- Products:
- 4 CO₂ → 4 × 2 = 8 O atoms
- 5 H₂O → 5 × 1 = 5 O atoms
→ Total = 8 + 5 = 13 O atoms

- Reactants:
O₂ provides oxygen in pairs. We need 13 oxygen atoms, but since O₂ is diatomic, we must have a coefficient that gives an even number. However, 13 is odd — so we can't use a whole number yet.

But wait! We need to balance oxygen. Since O₂ is diatomic, we'll adjust the coefficient of O₂ to provide 13 oxygen atoms.

But 13 is odd — so we need to avoid fractions initially. Let's double-check our work.

Wait — we have:
- 4 CO₂ → 8 O
- 5 H₂O → 5 O
→ Total O = 13

So we need 13 oxygen atoms from O₂. Since each O₂ molecule has 2 oxygen atoms, we need:

> 13/2 = 6.5 O₂ molecules

So we write:

> C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

But we don’t want fractional coefficients in final equations.

So, to eliminate the fraction, multiply all coefficients by 2:

> 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Now check if it’s balanced.

---

Step 3: Check the balance

| Element | Reactants | Products |
|--------|-----------|----------|
| C | 2 × 4 = 8 | 8 × 1 = 8 ✔ |
| H | 2 × 10 = 20 | 10 × 2 = 20 ✔ |
| O | 13 × 2 = 26 | (8 × 2) + (10 × 1) = 16 + 10 = 26 ✔ |

✔ All elements are balanced.

---

Final Balanced Equation:

> 2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

---

Answer Boxed:

$$
\boxed{2\,\text{C}_4\text{H}_{10} (g) + 13\,\text{O}_2 (g) \rightarrow 8\,\text{CO}_2 (g) + 10\,\text{H}_2\text{O} (g)}
$$

---

Explanation Summary:

1. Unbalanced equation: C₄H₁₀ + O₂ → CO₂ + H₂O
2. Balance C: 4 CO₂
3. Balance H: 5 H₂O
4. Balance O: 13/2 O₂ → then multiply entire equation by 2 to eliminate fraction
5. Final check: All atoms are balanced.

This is the balanced combustion reaction of butane.