Suppose you need 2.08 g of chlorine gas (Cl2) What volume at STP would you need to use?
To determine the volume of chlorine gas (Cl2) needed at STP (Standard Temperature and Pressure), we can use the ideal gas law:
PV = nRT
Where:
P = Pressure (in atmosphere, atm)
V = Volume (in liters, L)
n = Number of moles (in moles, mol)
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin, K)
At STP, the temperature is 273.15 K, and the pressure is 1 atmosphere (1 atm).
First, we need to determine the number of moles of chlorine gas (Cl2) using its molar mass. The molar mass of chlorine gas is 70.90 g/mol.
moles = mass / molar mass
moles = 2.08 g / 70.90 g/mol
moles ≈ 0.029 mol
Now, we can rearrange the ideal gas law equation to solve for the volume (V):
V = nRT / P
V = (0.029 mol) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm)
Simplifying the equation:
V ≈ 0.682 L
Therefore, you would need approximately 0.682 liters of chlorine gas (Cl2) at STP.
Use PV = nRT
n = grams/molar mass