Argon gas at standard temperature and pressure occupies 66 L. How many grams does this quantity of Argon weigh?
1 mol of Ar has a mass of 39.948 grams. 1 mol Ar at STP occupies 22.4 L. How many mols do you have in 66 L @ STP?
To find the weight of argon gas, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
In this case, we are given the volume (V) of the argon gas, which is 66 L.
Standard temperature and pressure (STP) is defined as 0 degrees Celsius (273.15 Kelvin) and 1 atmosphere (atm) of pressure.
The ideal gas constant (R) has a value of 0.0821 L·atm/(mol·K).
To find the number of moles (n), we need the pressure (P) and temperature (T). Since the problem states STP, P = 1 atm and T = 273.15 K.
Now we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Plugging in the values:
n = (1 atm) * (66 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
Calculating this equation will give us the number of moles of argon gas.
Once we have the number of moles, we can calculate the weight by multiplying it by the molar mass of argon.
The molar mass of argon is 39.95 g/mol.
Weight (g) = number of moles * molar mass