9of11 Part A Three gases (8.00 g of methane, CH, 18.0 g of ethane, C2H6, and an

Question

9of11 Part A Three gases (8.00 g of methane, CH, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3 Hs) were added to the same 10.0-L container. At 23.0 °C, the total pressure in the container is 4.60 atn. Calculate the partial pressure of each gas in the container Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane. View Available Hint(s) 0 ? atm Submit Previous Answers X Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

molar mass of methane = 16 g/mol then 8 gm methane = 8/16 = 0.5 mole

molar mass of ethane = 30 g/mol then 18 gm ethane = 18/30 = 0.6 mole

now calculate total mole of gas by using ideal gas equation

Ideal gas equation

PV = nRT             where, P = atm pressure= 4.60 atm,

V = volume in Liter = 10 L

n = number of mole = ?

R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,

T = Temperature in K = 230C = 273.15+ 23 = 296.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (4.60X 10)/(0.08205 X 296.15) = 1.9 mole

total mole of gas = 1.9 mole

mole of propane = total mole - mole of methane - mole of ethane

mole of propane = 1.9 - 0.5 -0.6 = 0.8 mole

now calulate mole fraction of each gas

mole fraction = no. of mole of insivisual gas / total mole

mole fraction of methane = 0.5 / 1.9 = 0.2632

mole fraction of ethane = 0.6 / 1.9 = 0.3158

mole fraction of propane = 0.8 / 1.9 = 0.421

now calculte partial pressure of each gas

partial pressure = mole fraction X  total pressure

partial pressure of methane = 0.2632 X 4.60 = 1.2 atm

partial pressure of ethane = 0.3158X 4.60 = 1.5 atm

partial pressure of propane = 0.421 X 4.60 = 1.9 atm

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