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tommyo0729 tommyo0729
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12 years ago

If 24.5 g of methane undergoes a combustion reaction with oxygen and methane is the limiting reagent, what is the maximum number of grams of water that can be produced?
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12 years ago
You have :
---------------------------------------------------------------------

  CH4  +  2 O2  --------> CO2  + 2 H2O  ..... [ Mole Basis ]

  16 CH4  + 64 O2  ------>  44 CO2  + 36 H2O .... [ Mass Basis }

  m sub CH4  =  24.5 g  CH4

   n sub CH4 = 24.5 / 16  = 1.53 moles CH4

  n sub O2 required  = ( 1.53 CH4 ) ( 2 O2 / 1 CH4 )  = 3.06 moles O2 required

  m sub O2 required = ( 3.06 ) ( 32 ) = 98.0 g O2 required

  n sub CO2  = ( 1.53 CH4 ) ( 1 CO2 / 1 CH4 )  = 1.53 moles CO2 produced

  m sub CO2 = ( 1.53 ) ( 44 )  = 37.4 g CO2  produced

  n sub H2O  = ( 1.53 CH4 ) ( 2 H2O / 1 CH4 )  = 3.06  moles H2O produced

  m sub H2O = ( 3.06 ) ( 18 ) = 55.1 g H2O produced

Now, what was your question ?
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