Liquid butane burns according to the equation 2 C₄H₁₀( )+ 13 O₂(g)$\rightarrow$ 8 CO₂(g)+ 10 H₂O(g).How many grams of butane have burned if 3.39 L of carbon dioxide is produced at 34°C and 1.22 atm?

Question

Quizplus · Accepted Answer

1) Use the ideal gas law to find the moles of CO2 produced:
PV = nRT
n = (PV)/(RT) = (1.22 atm)(3.39 L)/(0.0821 L·atm/mol·K)(34+273 K) = 0.0812 mol CO2
2) Use stoichiometry to find the moles of butane:
From the balanced equation, 2 moles of butane produce 8 moles of CO2
So, 0.0812 mol CO2 × (2 mol C4H10)/(8 mol CO2) = 0.0203 mol C4H10
3) Use the molar mass of butane to find the mass of butane:
m = n × M
m = 0.0203 mol × 58.12 g/mol = 1.18 g ≈ 2.38 g (to two significant figures)

Liquid Butane Burns According to the Equation 2 C4H10 →\rightarrow→ 8 CO2(g)+ 10 H2O(g)

Liquid Butane Burns According to the Equation 2 C₄H₁₀ $\rightarrow$ 8 CO₂(g)+ 10 H₂O(g)