What mass of propane gas is burned in a barbeque over a month at SATP conditions if 2.2 kL of gas is used?
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.