A particular natural gas consists, in mole percents, of 83.0% CH4 (methane), 11.

Question

A particular natural gas consists, in mole percents, of 83.0% CH4 (methane), 11.2% C2H6 (ethane), and 5.80% C3H8 (propane). A 385-L sample of this gas, measured at 25 C and 729 mmHg , is burned in an excess of oxygen gas.

How much heat, in kilojoules, is evolved in this combustion reaction?

Express your answer with the appropriate units.

The table shown here gives the enthalpy of combustion for three different hydrocarbon fuels to produce liquid water and gaseous carbon dioxide.

* The answer is not (15635.198 kJ/mol) or (16169.2 kJ).

Explanation / Answer

we know that

PV = nRT

given

V = 385

P = ( 729/760) atm

T = 298 K

so

PV = nRT

(729/760) x 385 = n x 0.0821 x 298

n = 15.09438

now

moles of CH4 = 0.83 x 15.09438

moles of Ch4 = 12.528

moles of C2H6 = 0.112 x 15.09438

moles of C2H6 = 1.69

now

moles of C3H8 = 0.058 x 15.09438

moles of C3H8 = 0.8755

now

total heat = heat from CH4 + heat from C2H6 + heat from C3H8

also

heat = moles x dH

so

total heat = -( 12.528 x 890.3 ) + ( 1.69 x 1559.7) + ( 0.8755 x 2219.1)

total heat = -15732.33

so

the heat is -15732.33 kJ

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