Question

A natural gas has the following composition: CH4 (methane) = 87 %, C2H6 (ethane) = 12 %, and C3H8 (propane) = 1 %.

a.) What is the composition in weight percent?

b.) What is the composition in volume percent?

c.) How many cubic meters will be occupied by 80.0 kg of the gas at 9 °C and 600 kPa?

d.) What is the density of the gas in kilograms per cubic meter at standard conditions?

e.) What is the specific gravity at 9 °C and 600 kPa referred to air at standard conditions?

Answer 1

a) Molar Mass of CH4 = 16g/mol

Molar mass of C2H6 = 30g/mol

Molar mass of propane = 44g/mol

For 100moles

87 moles of CH4 ; mass 87 × 16g = 1392g

12 moles of C2H6; mass 12 × 30g = 360g

1 mole of C2H6 ; mass 1× 44g = 44g

Total mass = 1796g

Therefore,

Weight ℅ of CH4 = (1392g/1796g)×100 = 77.51℅

Weight % of C2H6 = ( 360g/1796g)×100 = 20.04%

Weight ℅ of C2H6 = (44g/1796g)×100= 2.45℅

b) Volume ℅ of CH4 = 87℅

Volume % of C2H6 = 12℅

Volume ℅ of C2H6 = 1℅

C) Tota moles = (100moles/1.796Kg)×80kg = 4454.34mols

PV = nRT

V = nRT/P

= 4454.34 mol × 0.082057( L atm/mol K) × 282K/5.922atm

= 17405L

= 17.41m^3

d) Density = Mass/Volume

For 100 mole , mass = 1.796g

V = nRT/P

= 100mol × 0.082057(L atm/mol K)×273K/1atm

= 2240L

= 2.240m^3

Density = 1.796kg/2.240m^3 = 0.802kg/m^3

e) specific gravity is relative Density

Spgr = Density at 9℃ and 600kPa / Density of air at std condition

= 4.596(kg/m^3)/1.204(kg/m^3)

= 3.82