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If 8.50 L of natural gas, which is essentially methane (CH4), undergoes complete combustion at 730 mm Hg and 20 degrees C, how many grams of each product are formed?

Grams of CO2=______
Grams of H2O=______

1 Answer

6 votes

mass CO₂ : 14.96 g

mass H₂O : 12.24 g

Further explanation

Reaction

CH₄+2O₂⇒CO₂+2H₂O

  • mol CH₄

V=8.5 L

P=730 mmHg=0,961 atm

T=20+273=293 K


\tt n=(PV)/(RT)=(0.961* 8.5)/(0.082* 293)=0.34

  • mass CO₂(MW=44 g/mol)

mol CO₂= mol CH₄ = 0.34


\tt 0.34* 44=14.96~g

  • mass H₂O (MW=18 g/mol)

mol H₂O= 2 x mol CH₄ = 2 x 0.34 = 0.68


\tt 0.68* 18=12.24~g

User James Shaw
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