Answer: 0.67 L of fluorine gas would be produced when 0.67 L of HF reacts with excess O2
The balanced chemical equation for the reaction between hydrogen fluoride (HF) and oxygen (O2) is:
2 HF + O2 -> 2 F2 + 2 H2O
From this equation, we can see that two moles of HF react with one mole of O2 to produce two moles of fluorine gas (F2).
To calculate the amount of F2 produced, we need to determine the number of moles of HF present in 0.67 L of HF. We can use the ideal gas equation, which states that:
PV = nRT
where
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = gas constant
T = temperature of the gas.
Assuming standard temperature and pressure (STP) remains same, we would use the molar volume of a gas at STP, which is 22.4 L/mol, to convert the volume of HF to moles:
0.67 L HF x (1 mol HF / 22.4 L HF) = 0.03 mol HF
Since the reaction uses two moles of HF to produce two moles of F2, the number of moles of F2 produced is also 0.03 mol.
To convert moles of F2 to volume at STP, we can again use the molar volume of a gas:
0.03 mol F2 x (22.4 L F2 / 1 mol F2) = 0.67 L F2
Thus, 0.67 L of fluorine gas can be produced when 0.67 L of HF reacts with excess O2.