1040 grams of chlorine gas at STP would have what volume?
To find the volume of chlorine gas at STP (Standard Temperature and Pressure) given its mass, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (at STP, the pressure is 1 atmosphere)
V = Volume (what we're trying to find)
n = Number of moles of gas
R = Ideal gas constant (0.0821 L.atm/(mol.K))
T = Temperature (at STP, the temperature is 273.15 K)
First, we need to determine the number of moles of chlorine gas using its mass and molar mass. The molar mass of chlorine (Cl2) is 70.906 g/mol.
Number of moles (n) = Mass (in grams) / Molar mass
n = 1040 g / 70.906 g/mol
Next, we can substitute the known values into the ideal gas law equation and solve for V:
1 atm * V = (1040 g / 70.906 g/mol) * (0.0821 L.atm/(mol.K)) * 273.15 K
Now, let's calculate the volume:
V = (1040 g / 70.906 g/mol) * (0.0821 L.atm/(mol.K)) * 273.15 K / 1 atm
By performing the calculations, we get:
V ≈ 383.85 L
Therefore, 1040 grams of chlorine gas at STP would have a volume of approximately 383.85 liters.