Consider the following balanced chemical equation for the combustion of propane.C3H8 + 5O2 ⟶ 3CO2 + 4H2OWhat volume of oxygen gas at 13.06 oC and 80.64 atm is needed for the complete combustion of 70.55 g of propane.
Question
Consider the following balanced chemical equation for the combustion of propane.C3H8 + 5O2 ⟶ 3CO2 + 4H2OWhat volume of oxygen gas at 13.06 oC and 80.64 atm is needed for the complete combustion of 70.55 g of propane.
Solution
To solve this problem, we need to follow these steps:
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Convert the mass of propane (C3H8) to moles. The molar mass of propane is approximately 44.1 g/mol. So, 70.55 g of propane is equivalent to 70.55 g / 44.1 g/mol = 1.6 moles.
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According to the balanced chemical equation, one mole of propane requires five moles of oxygen (O2) for complete combustion. Therefore, 1.6 moles of propane will require 1.6 * 5 = 8 moles of oxygen.
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Use the ideal gas law to find the volume of oxygen. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, convert the temperature from Celsius to Kelvin. T(K) = 13.06°C + 273.15 = 286.21 K.
The ideal gas constant (R) when pressure is in atm and volume is in liters is 0.0821 L.atm/(mol.K).
Rearrange the ideal gas law to solve for V: V = nRT / P.
Substitute the known values into the equation: V = (8 moles * 0.0821 L.atm/(mol.K) * 286.21 K) / 80.64 atm = 23.2 liters.
So, 23.2 liters of oxygen gas at 13.06 °C and 80.64 atm is needed for the complete combustion of 70.55 g of propane.
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