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=

mass CO₂ : 14.96 g

mass H₂O : 12.24 g

Reaction

CH₄+2O₂⇒CO₂+2H₂O

V=8.5 L

P=730 mmHg=0,961 atm

T=20+273=293 K

mol CO₂= mol CH₄ = 0.34

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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

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