To solve this problem, we first need to understand the combustion reaction of pentane (C5H12). The balanced chemical equation for the combustion of pentane is:
C5H12 + 8 O2 → 5 CO2 + 6 H2O
From this equation, we can see that 1 mole of pentane reacts with 8 moles of oxygen gas to produce 5 moles of carbon dioxide and 6 moles of water vapor. The reaction yields a total of 11 moles of gaseous products (5 moles of CO2 and 6 moles of H2O).
Next, we need to calculate the number of moles of pentane and oxygen available for the reaction:
Molar mass of pentane (C5H12):
- C: 12.01 g/mol × 5 = 60.05 g/mol
- H: 1.008 g/mol × 12 = 12.096 g/mol
- Total = 60.05 + 12.096 = 72.146 g/mol
Moles of pentane:
9.86 g C5H12 × (1 mol / 72.146 g) ≈ 0.137 mol
Molar mass of oxygen (O2): 32.00 g/mol
Moles of oxygen:
16.0 g O2 × (1 mol / 32.00 g) = 0.500 mol
Now, we determine which reactant is the limiting reagent. The stoichiometry of the reaction requires 8 moles of O2 for every mole of pentane:
0.137 mol C5H12 × 8 mol O2 / 1 mol C5H12 = 1.096 mol O2 needed
Since we only have 0.500 moles of O2, the oxygen is the limiting reagent. Now, we will use the moles of O2 to determine the moles of gaseous products produced:
Using the reaction coefficients, 0.500 mol O2 produces:
0.500 mol O2 × (5 mol CO2 / 8 mol O2) = 0.3125 mol CO2
0.500 mol O2 × (6 mol H2O / 8 mol O2) = 0.375 mol H2O
Total moles of gaseous products = 0.3125 + 0.375 = 0.6875 mol
Finally, we can find the pressure using the Ideal Gas Law: PV = nRT, where:
- P = pressure (in atm)
- V = volume (in liters) = 3.44 L
- n = number of moles of gas = 0.6875 mol
- R = ideal gas constant = 0.0821 L·atm/(K·mol)
- T = temperature in Kelvin (assuming room temperature ~ 25°C = 298 K)
Rearranging the equation gives us:
P = nRT / V
Plugging in our values:
P = (0.6875 mol) × (0.0821 L·atm/(K·mol)) × (298 K) / (3.44 L)
P ≈ 5.09 atm
Therefore, the pressure inside the vessel after the combustion of pentane would be approximately 5.09 atm.