Question

A fuel gas is known to contain methane, ethane, and carbon monoxide. A sample of the gas is charged into an initially evacuated 2.000-liter vessel at 25°C and 2323 mm Hg absolute. The vessel is weighed before and after being charged, and the mass difference is found to be 4.929 g. Next, the higher heating value of the gas is determined in a calorimeter to be 841.9 kJ/mol. Calculate the molar composition of the fuel gas. Calculate the lower heating value of the gas. State in your own words the physical significance of the calculated value.

Answer 1

To calculate the molar composition of the fuel gas, determine the moles of each component and divide by the total moles. The lower heating value of the gas can be found by subtracting the latent heat of water vapor from the higher heating value. The lower heating value represents the usable heat released from the combustion process.

Step-by-step explanation:

The molar composition of the fuel gas can be calculated by comparing the mass difference of the gas sample in the vessel to the total mass difference recorded. To calculate the molar composition, you would first determine the moles of methane, ethane, and carbon monoxide in the gas sample. Then, divide each mole by the total moles to find the molar composition.

The lower heating value of the gas can be calculated by subtracting the latent heat of water vapor from the higher heating value. The lower heating value represents the heat released when the water vapor in the products condenses to liquid.

The physical significance of the calculated lower heating value is that it gives an indication of the amount of heat that can be utilized from the combustion of the fuel gas. It represents the maximum amount of heat that can be obtained when the water vapor in the products is condensed.

Answer 2

Step-by-step explanation:

Solution

Let n1, n2, n3 be the moles of ethane, CO and methane respectively, then total number of moles n,

n =n1+n2-n3

Then

n = PV/RT

substituting the values gives

n= 0.25 moles

total wt of gas mixture = 4.929 g

Molecular wt of mixture MWmix = 4.929.0.25 = 19.716 g/ mol

if M1 , M2, M3 be the molecular wt of ethane, CO and methane

M1 = 30.07, M2 = 28.01, M3 = 28.01

But

MWmix = y1 * M1 + y2 * M2 + y3* M3

y1 + y2 + y3 = 1

where y1 = n1/n, y2 = n2/n, y3=n3/n

Solving gives y1 (moles of ethane) =0 .103

Solving gives y2 (moles of CO) = 0.193

Solving gives y1 (moles of Methane) =0 .704

Therefore gas mixture is

ethane = 10.3 %

CO = 19.3 %

Methane = 70.4 %