Asked by Anonymous

Consider the following reaction.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) ÄH = -891 kJ
Calculate the enthalpy change for each of the following cases.
(a) 5.00 g methane is burned in excess oxygen.

15 years ago

Answered by DrBob222

You omitted the arrow. We can't separate products from reactants without an arrow.

-891 kJ x (5.00/16) = ?? kJ.

15 years ago

Answered by Anonymous

5gCH4/16gCH4= 0.3125moles CH4
(0.3125molesCH4)x(891kJ)=278kJ
Enthalpy change = 278kJ

14 years ago

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