To solve this problem, we need to first convert the volume of propane gas (2.2 kL) to mass using the ideal gas law equation:
PV = nRT
where P is the pressure (in this case, SATP conditions at 1 atm), V is the volume of the gas, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Assuming the temperature is around 25 degrees Celsius, we can use R = 0.0821 L.atm/mol.K.
Rearranging the equation to solve for n:
n = PV / RT
n = (1 atm)(2.2 kL) / (0.0821 L.atm/mol.K * 298 K)
n = 88.2 mol
Now we need to convert moles to mass using the molar mass of propane (C3H8). The molar mass of propane is approximately 44.11 g/mol.
Mass of propane gas = 88.2 mol * 44.11 g/mol
Mass of propane gas = 3,892.10 g
Therefore, approximately 3,892.10 grams (3.9 kg) of propane gas is burned in a barbecue over a month at SATP conditions if 2.2 kL of gas is used.
What mass of propane gas is burned in a barbeque over a month at SATP conditions if 2.2 kL of gas is used?
1 answer