a sample of natural gas at stp is 78% ch4 and 22% c2h6 by volume. how many moles of each gas are presented in 2 L of the mixture.assume ideal gas behavior
To find the number of moles of each gas in the mixture, we need to use the ideal gas equation:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume (2 L in this case)
n = moles of gas
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (at STP, it is 273 K)
Let's start by calculating the number of moles of CH4 (methane) in 2 L of the mixture.
Since the volume percentage of CH4 is given, we'll use it to find the volume of CH4 in the mixture:
Volume of CH4 = 78% * 2 L = 1.56 L
Now, we can calculate the moles of CH4 by dividing its volume by the molar volume:
n(CH4) = Volume of CH4 / Molar volume
The molar volume at STP is 22.4 L/mol.
n(CH4) = 1.56 L / 22.4 L/mol
n(CH4) ≈ 0.07 mol
So, there are approximately 0.07 moles of CH4 in 2 L of the mixture.
Next, let's calculate the number of moles of C2H6 (ethane):
Volume of C2H6 = 22% * 2 L = 0.44 L
n(C2H6) = Volume of C2H6 / Molar volume
n(C2H6) = 0.44 L / 22.4 L/mol
n(C2H6) ≈ 0.02 mol
Therefore, there are approximately 0.02 moles of C2H6 in 2 L of the mixture.
In summary, there are approximately 0.07 moles of CH4 and 0.02 moles of C2H6 in 2 L of the natural gas mixture.