Question

balance the chemical equation below using the smallest possible whole number stoichiometric coefficients. \\( \ce{ch3ch3(g) + o2(g) -> co2(g) + h2o(g)} \\)

Explanation:

Step1: Identify the reactants and products

The reaction is \( \text{CH}_3\text{CH}_3(g) + \text{O}_2(g)
ightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \). Reactants: \( \text{C}_2\text{H}_6 \) (ethane, since \( \text{CH}_3\text{CH}_3 \) is ethane) and \( \text{O}_2 \). Products: \( \text{CO}_2 \) and \( \text{H}_2\text{O} \).

Step2: Balance Carbon atoms

Ethane (\( \text{C}_2\text{H}_6 \)) has 2 C atoms. So, we need 2 \( \text{CO}_2 \) molecules to balance C. Now the equation becomes: \( \text{CH}_3\text{CH}_3(g) + \text{O}_2(g)
ightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(g) \)

Step3: Balance Hydrogen atoms

Ethane has 6 H atoms. Each \( \text{H}_2\text{O} \) has 2 H atoms. So, number of \( \text{H}_2\text{O} \) molecules needed: \( \frac{6}{2} = 3 \). Now the equation: \( \text{CH}_3\text{CH}_3(g) + \text{O}_2(g)
ightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \)

Step4: Balance Oxygen atoms

On the product side: \( 2 \times 2 + 3 \times 1 = 4 + 3 = 7 \) O atoms. But \( \text{O}_2 \) is diatomic, so we need \( \frac{7}{2} \) \( \text{O}_2 \), but we need whole numbers. Multiply all coefficients by 2 to eliminate the fraction.

Multiply each coefficient by 2:

• \( \text{CH}_3\text{CH}_3 \): \( 2 \times 1 = 2 \)
• \( \text{O}_2 \): \( 2 \times \frac{7}{2} = 7 \)
• \( \text{CO}_2 \): \( 2 \times 2 = 4 \)
• \( \text{H}_2\text{O} \): \( 2 \times 3 = 6 \)

Now the balanced equation: \( 2\text{CH}_3\text{CH}_3(g) + 7\text{O}_2(g)
ightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \)

Answer:

The balanced chemical equation is \( \boldsymbol{2}\text{CH}_3\text{CH}_3(g) + \boldsymbol{7}\text{O}_2(g)
ightarrow \boldsymbol{4}\text{CO}_2(g) + \boldsymbol{6}\text{H}_2\text{O}(g) \)