Assuming no change in temperature and pressure, calculate the volume of O₂ (in liters) required for the complete combustion of 14.9 L of butane (C₄H₁₀):
2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)
2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)
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SOLUTION:
Based on the balanced chemical equation, 2 liters of C₄H₁₀ is stoichiometrically equivalent to 13 liters of O₂.
[tex]\begin{aligned} \text{volume of} \: \text{O}_2 & = \text{14.9 L} \: \text{C}_4\text{H}_{10} \times \frac{\text{13 L} \: \text{O}_2}{\text{2 L} \: \text{C}_4\text{H}_{10}} \\ & = \boxed{\text{96.85 L}} \end{aligned}[/tex]
Hence, 96.85 L of O₂ is required for the complete combustion of 14.9 L of C₄H₁₀.
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