Given:
For butane (C₄H₁₀)
[tex]V = \text{285 L}[/tex]
[tex]P = \text{1.04 atm}[/tex]
[tex]T = \text{25°C + 273 = 298 K}[/tex]
Required:
mass of CO₂
Solution:
Step 1: Calculate the number of moles of butane.
[tex]PV = nRT[/tex]
[tex]n = \frac{PV}{RT}[/tex]
[tex]n = \frac{\text{(1.04 atm)(285 L)}}{(0.0821 \: \frac{\text{L • atm}}{\text{mol • K}})(\text{298 K})}[/tex]
[tex]n = \text{12.115 mol}[/tex]
Step 2: Calculate the molar mass of CO₂.
molar mass = (12.01 g/mol × 1) + (16.00 g/mol × 2)
molar mass = 44.01 g/mol
Step 3: Determine the mole ratio needed.
mole ratio = 2 mol C₄H₁₀ : 8 mol CO₂
Step 4: Calculate the mass of CO₂ produced using the mole ratio.
[tex]\text{mass of CO₂ = 12.115 mol C₄H₁₀} × \frac{\text{8 mol CO₂}}{\text{2 mol C₄H₁₀}} × \frac{\text{44.01 g CO₂}}{\text{1 mol CO₂}}[/tex]
[tex]\boxed{\text{mass of CO₂ = 2130 g}}[/tex]
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