Question

balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.
ch₃(ch₂)₃ch₃(l) + o₂(g) → co₂(g) + h₂o(g)

Explanation:

Step1: Identify the compound formula

The given compound is \( \text{CH}_3(\text{CH}_2)_3\text{CH}_3 \) (also known as pentane, \( \text{C}_5\text{H}_{12} \)). So the unbalanced equation is \( \text{C}_5\text{H}_{12}(l) + \text{O}_2(g)
ightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \).

Step2: Balance Carbon atoms

There are 5 C atoms in \( \text{C}_5\text{H}_{12} \), so we put a coefficient of 5 in front of \( \text{CO}_2 \):
\( \text{C}_5\text{H}_{12}(l) + \text{O}_2(g)
ightarrow 5\text{CO}_2(g) + \text{H}_2\text{O}(g) \)

Step3: Balance Hydrogen atoms

There are 12 H atoms in \( \text{C}_5\text{H}_{12} \). Since each \( \text{H}_2\text{O} \) has 2 H atoms, we need \( \frac{12}{2} = 6 \) \( \text{H}_2\text{O} \):
\( \text{C}_5\text{H}_{12}(l) + \text{O}_2(g)
ightarrow 5\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \)

Step4: Balance Oxygen atoms

On the right side, total O atoms are \( (5 \times 2) + (6 \times 1) = 10 + 6 = 16 \). Each \( \text{O}_2 \) has 2 O atoms, so we need \( \frac{16}{2} = 8 \) \( \text{O}_2 \):
\( \text{C}_5\text{H}_{12}(l) + 8\text{O}_2(g)
ightarrow 5\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \)

Answer:

\( \text{CH}_3(\text{CH}_2)_3\text{CH}_3(l) + 8\text{O}_2(g)
ightarrow 5\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \) (or with the formula written as \( \text{C}_5\text{H}_{12}(l) + 8\text{O}_2(g)
ightarrow 5\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \))