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?
L CH4 = 1.60 x 0.78 = ??
L C2H6 = 1.60 x 0.22 = ??
moles = L/22.4
To find the number of moles of each gas present in 1.60L of the mixture, you need to use the ideal gas law equation.
First, let's calculate the number of moles of CH4 (methane):
1. Calculate the volume of CH4 using the partial pressure:
Volume of CH4 = 78% * 1.60 L = 1.25 L
2. Use the ideal gas law equation to find the number of moles of CH4:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L*atm/(mol*K))
T = temperature in Kelvin (at STP, it is 273.15 K)
Rearrange the equation to solve for n:
n = PV / RT
n(CH4) = (1 atm) * (1.25 L) / (0.0821 L*atm/(mol*K) * 273.15 K)
n(CH4) ≈ 0.06 moles
Next, let's calculate the number of moles of C2H6 (ethane):
1. Calculate the volume of C2H6 using the partial pressure:
Volume of C2H6 = 22% * 1.60 L = 0.35 L
2. Use the ideal gas law equation to find the number of moles of C2H6:
n(C2H6) = (1 atm) * (0.35 L) / (0.0821 L*atm/(mol*K) * 273.15 K)
n(C2H6) ≈ 0.02 moles
Therefore, in 1.60L of the mixture, there are approximately 0.06 moles of CH4 (methane) and 0.02 moles of C2H6 (ethane).