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Calculate the volume of air, contained 21% of oxygen by volume required for the complete combustion of 10 L of ethylene under similar conditions |
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Answer» Solution :Combustion of ethylene is given by the equation `C_(2)H_(4)+3O_(2) to 2CO_(2)+2H_(2)O` ONE mole of `C_(2)H_(4)="3 moles of "O_2` one volume of `C_(2)H_(4)="3 volumes of "O_(2)` 10mL of `C_(2)H_(4)=?` The volume of `O_(2)` required for the combustion of 10 L of `C_(2)H_(4)=10 XX 3=30L` But AIR contains only 21% `O_(2)` by volume 21 L of `O_(2)` =100 of air 30 L of `O_(2)=?` The volume of air required for the combustion `=(30)/(21) xx 100=142.86L` |
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