total mass of water formed NEED ANSWER FOR PART B

Question

Natural gas is a mixture of hydrocarbons, primarily methane (CH₄) and ethane (C₂H₆). A typical mixture might have X_methane= 0.915 and X_ethane = 0.085.    (R= 0.0821 L atm/mol K)

a.) What are the partial pressures of the two gases in a 15.00 L container of natural gas at 20.°C and 1.44 atm?

Given mole fraction of methane , X_methane = 0.915

and mole fraction of ethane X_ethane = 0.085

Total pressure, P = 1.44 atm

Partial pressure of methane, p _metahne = total pressure * mole fraction of methane

p_metahne = P * X_methane

= 1.44 atm * 0.915 = 1.32 atm

Partial pressure of ethane, p _etahne = total pressure * mole fraction of ethane

p_etahne = P * X_ethane

= 1.44 atm * 0.085 = 0.12 atm

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

Answer 1

Let's look at the equations for combustion of each gas:

CH₄ + 2O₂ ==> CO₂ + 2H₂O ... complete combustion of methane

2C₂H₆ + 7O₂ ==> 4CO₂ + 6H₂O ... complete combustion of ethane

moles CH₄ present = PV/RT = (1.32 atm)(15 L)/(0.0821)(293) = 0.823 moles

mass H₂O produced = 0.823 mol CH₄ x 2 mol H₂O/mol CH₄ x 18 g mol = 29.6 g H₂O

moles C₂H₆ present = PV/RT = (0.12 atm)(15 L)/(0.0821)(293) = 0.0749 moles

mass H₂O produced = 0.0749 mol C₂H₆ x 6 mol H₂O/2 mol C₂H₆ x 18 g/mol = 4.05 g H₂O

Total mass H₂O produced = 29.6 g + 4.05 g = 33.6 g H₂O