A container contains 3.2 g of O2 gas. The volume of the gas at STP is
A. 2.24L
B. 224.0L
C. 0.224L
D. None
To determine the volume of the gas at STP, we need to use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
At STP, the pressure (P) is 1 atm, and the temperature (T) is 273 K.
First, we need to determine the number of moles of O2 gas:
n = mass / molar mass
The molar mass of O2 gas is 32 g/mol (2 * 16 g/mol).
n = 3.2 g / 32 g/mol
n = 0.1 mol
Now we can rearrange the ideal gas law equation to solve for the volume (V):
V = nRT / P
Substituting the known values:
V = (0.1 mol * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm
V = 2.2413 L
So the volume of the gas at STP is approximately 2.2413 L.
Therefore, the correct answer is A. 2.24 L.
3.2 g = ?mols O2.
mols O2 = 3.2/molar mass O2 = ?
Then you know that 1 mol of a gas occupies 22.4 L at STP