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Thermodynamics

Question
CBSEENCH11006201

How many grams of methane and volume of oxygen at N.T.P. would be required to produce 445.15 kJ of heat in the following reaction?

CH subscript 4 left parenthesis straight g right parenthesis space plus space 2 straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space space CO subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon
                                                          increment straight H space equals space minus 890.3 space k J

                                       

Solution

The thermochemical reaction is,
stack CH subscript 4 left parenthesis straight g right parenthesis with stack 12 plus 4 with equals 16 space straight g below below space plus stack space 2 straight O subscript 2 left parenthesis straight g right parenthesis space with stack 2 space cross times space 22.4 with equals space 44.8 space space liltrr below below space space space rightwards arrow space space space space space stack CO subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 straight O space left parenthesis straight l right parenthesis with increment straight H space equals space minus 890.3 space kJ below

Step I.
Amount of methane required:
890.3 kJ of heat is produced from methane = 16 g
therefore 445.15 kJ of heat is produced from methane
                                        equals space fraction numerator 16 over denominator 890.3 end fraction cross times 456.15
equals space 8 straight g

Step II.
Volume of oxygen required:
890.3 kJ heat is evolved from oxygen = 44.8 litres
therefore 445.15 kJ of heat is evolved from oxygen
                                   equals space fraction numerator 44.8 over denominator 890.3 end fraction cross times 445.15
equals space 22.4 space litres.