A sample of natural gas at STP is 78% CH4 and 22% C2H6 by volume. How many moles of each gas are present in 1.60L of the mixture? Assume ideal gas behavior
Wouldn't that be 0.22 x 1.60 L and
0.78 x 1.60 L to obtain volume, then convert volume at STP to moles?
To determine the number of moles of each gas in the mixture, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since the sample of natural gas is at Standard Temperature and Pressure (STP), we can assume that the pressure (P) is 1 atmosphere (atm) and the temperature (T) is 273 Kelvin (K).
First, let's calculate the number of moles of CH4 (methane) in the mixture:
1. Calculate the volume occupied by CH4:
Using the volume percentage, we can calculate the volume of CH4 in the mixture.
Volume of CH4 = volume of mixture * Volume percentage of CH4
Volume of CH4 = 1.60 L * 0.78
Volume of CH4 = 1.248 L
2. Convert the volume of CH4 to moles:
To convert the volume to moles, we need to use the ideal gas law equation and solve for n (moles).
PV = nRT
n = PV / RT
n(CH4) = (1 atm * 1.248 L) / (0.0821 atm L/mol K * 273 K)
n(CH4) = 0.0591 moles
Next, let's calculate the number of moles of C2H6 (ethane) in the mixture:
1. Calculate the volume occupied by C2H6:
Volume of C2H6 = volume of mixture * volume percentage of C2H6
Volume of C2H6 = 1.60 L * 0.22
Volume of C2H6 = 0.352 L
2. Convert the volume of C2H6 to moles:
n(C2H6) = (1 atm * 0.352 L) / (0.0821 atm L/mol K * 273 K)
n(C2H6) = 0.0167 moles
Therefore, in 1.60 L of the natural gas mixture, there are approximately 0.0591 moles of CH4 and 0.0167 moles of C2H6.