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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=______

Respuesta :

ardni313

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

[tex]\tt n=\dfrac{PV}{RT}=\dfrac{0.961\times 8.5}{0.082\times 293}=0.34[/tex]

  • mass CO₂(MW=44 g/mol)

mol CO₂= mol CH₄ = 0.34

[tex]\tt 0.34\times 44=14.96~g[/tex]

  • mass H₂O (MW=18 g/mol)

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

[tex]\tt 0.68\times 18=12.24~g[/tex]

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