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Question

Q17. What volume of air (containing20%O2 by volume) will be required to burn completely 10cm3each of methane and acetylene?


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Solution

  • The balanced equation for CH4:

CH4methane1vol(g)+2O2oxygen2vol(g)CO2carbondioxide1vol(g)+2H2Owater2vol(l)

  • The balanced equation for C2H2:

2C2H2acetylene2vol(g)+5O2oxygen5vol(g)4CO2carbondioxide4vol(g)+2H2O(l)water2vol

  • Calculate the volume of O2 for CH4

From the above equation it is clear that1 volume of CH4requires 2vol of O2

Therefore, 10cm3of CH4 will require =10×2=20cm3 of O2

  • Calculate the volume of O2 for C2H2

Also 2vol of C2H2 require 5vol of O2

Therefore 10cm3 will require O2for complete combustion=52×10=25cm3

Now total O2 required=20+25=45cm3

Since air contains 20%of O2 by volume

  • Thus the volume of air containing 45cm3of O2is 225cm3.


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