Question

Natural gas is a mixture of hydrocarbons, primarily methane (CH4) and ethane (C2H6). A typical mixture might have Xmethane=0.915 and Xethane=0.085. What are the partial pressures of the two gases in a 13.80-L container of natural gas at 20.°C and 1.40 atm?

Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?

Answer 1

The partial pressure can be calculated from the molar fraction and total pressure:

p_(CH_4 )=p_0·x_(CH_4 )=1.40 atm·0.915=1.281 atm
p_(C_2 H_6 )=p_0·x_(C_2 H_6 )=1.40 atm·0.085=0.119 atm.
Considering the combustion of methane and ethane, let’s write the reactions equations:

CH_4+〖2O〗_2→CO_2+2H_2 O
〖2C〗_2 H_6+7O_2→4CO_2+6H_2 O.
As one can see, one mole of methane produces 2 moles of water, and 2 moles of ethane produce 6 moles of water.

n_(CH_4 )=n_(H_2 O-1)/2
n_(C_2 H_6 )=n_(H_2 O-2)/3.
Let’s calculate the number of the moles of methane and ethane, using ideal gas law:

n_tot= pV/RT=(1.40atm·13.80L)/((0.08206 L·atm·mol^(-1)·K^(-1))(293.15 K) )=0.803 mol.
Thus, the number of the moles of methane:

n_(CH_4 )=n_tot·x_(CH_4 )= 0.803 (mol)·0.915=0.735 mol.
The number of the moles of ethane:

n_(C_2 H_6 )=n_tot·x_(C_2 H_6 )= 0.803 (mol)·0.085=0.068 mol.
The total mass of water is:

n_(H_2 O)= n_(H_2 O-1)+n_(H_2 O-2)=2n_(CH_4 )+3n_(C_2 H_6 ).

n_(H_2 O)=2·0.735+3·0.068=1.675 mol

The mass of water produced due to the combustion is:

m_(H_2 O)=n_(H_2 O)·M_(H_2 O)=1.675 mol·18.01528 g mol^(-1)=30.17 g

Answer: The partial pressures of methane and ethane are 1.281 atm and 0.119 atm, respectively. 30.17 g of water is formed in the case of complete combustion of both gases.