How many moles of oxygen gas (O2) are in 448L at STP (273K and 1 atm of pressure)?
Use PV = nRT
To find out how many moles of oxygen gas (O2) are in 448L at STP (273K and 1 atm of pressure), we can use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
Since we have the volume (V), pressure (P), and temperature (T), we can rearrange the equation to solve for the number of moles (n):
n = PV / RT
First, convert the volume to liters:
448L = 448 liters
Next, substitute the given values into the equation:
P = 1 atm
V = 448 L
R = 0.0821 L.atm/mol.K
T = 273 K
Plugging these values into the equation:
n = (1 atm * 448 L) / (0.0821 L.atm/mol.K * 273 K)
Simplifying the equation gives:
n = 5.52 moles
Therefore, there are approximately 5.52 moles of oxygen gas (O2) in 448L at STP.