What volume of oxygen gas measured at 0°C and 1 atm is needed to burn completely 1L of propane gas measured under same conditions?
C3H8 + 5O2>>>3CO2 + 4H2O
So the law of volumes: 5 liters of O2
To determine the volume of oxygen gas needed to burn 1L of propane gas, we need to look at the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the balanced equation, we can see that 1 mole of propane gas (C3H8) reacts with 5 moles of oxygen gas (O2) to produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O).
To calculate the volume of oxygen gas, we need to convert the volume of propane gas (1L) to moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (1 atm)
V = volume (1L)
n = moles of gas (unknown)
R = ideal gas constant (0.0821 L∙atm/(mol∙K))
T = temperature (0°C + 273.15 = 273.15K)
Using the ideal gas law equation, we can rearrange it to solve for n:
n = PV / RT
n = (1 atm) * (1L) / (0.0821 L∙atm/(mol∙K)) * (273.15K)
n ≈ 0.0445 moles of propane gas
Now, we can calculate the moles of oxygen gas needed by multiplying the moles of propane gas by the stoichiometric ratio from the balanced equation:
moles of O2 = 5 * (moles of C3H8)
moles of O2 = 5 * 0.0445
moles of O2 ≈ 0.2225 moles of oxygen gas
To convert the moles of oxygen gas to volume, we can use the ideal gas law equation rearranged to solve for V:
V = nRT / P
V = (0.2225 moles) * (0.0821 L∙atm/(mol∙K)) * (273.15K) / (1 atm)
V ≈ 4.65L
Therefore, approximately 4.65L of oxygen gas, when measured at 0°C and 1 atm, is needed to completely burn 1L of propane gas under the same conditions.