Question

How much $CO_{2}$ is emitted in the atmosphere if 150 g of butane is oxidized fully into water and Carbon-Dioxide?

• A. \( 455g \)
• B. \( 400g \)
• C. \( 350g \)
• D. \( 525g \)

Hint:

For rapid calculation, use the mass ratio directly: \( \text{Mass of } CO_{2} = \text{Initial Mass} \times \frac{4 \times M_{CO_{2}}}{M_{Butane}} \). Substituting: \( 150 \times \frac{176}{58} \approx 455.17~g \).

Answer 1

The Correct Option is A

Concept: The complete oxidation (combustion) of butane (\( C_{4}H_{10} \)) follows a stoichiometric relationship where carbon atoms in the fuel are converted entirely into \( CO_{2} \). [cite: 20]
• Balanced Reaction: \( 2C_{4}H_{10} + 13O_{2} \rightarrow 8CO_{2} + 10H_{2}O \) [cite: 20]
• This implies 1 mole of Butane produces 4 moles of \( CO_{2} \). [cite: 20]

Step 1: Calculate the molar masses.

• Molar mass of Butane (\( C_{4}H_{10} \)) = \( (4 \times 12) + (10 \times 1) = 58~g/mol \) [cite: 20]
• Molar mass of \( CO_{2} \) = \( 12 + (2 \times 16) = 44~g/mol \) [cite: 20]

Step 2: Determine the moles of Butane.
Given mass of butane = 150 g. [cite: 20] \[ n_{butane} = \frac{150~g}{58~g/mol} \approx 2.586~mol \]

Step 3: Calculate the mass of \( CO_{2} \) produced.
Since 1 mole of butane yields 4 moles of \( CO_{2} \): \[ n_{CO_{2}} = 2.586 \times 4 = 10.344~mol \] \[ \text{Mass of } CO_{2} = 10.344~mol \times 44~g/mol \approx 455.14~g \] Rounding to the nearest whole number gives 455g. [cite: 21]