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What volume of air (in `m^(3)`) is needed for the combustion of `1m^(3)` of a gas having the following composition in percentage volume : `2%` of `C_(2)H_(2), 8%` of `CO, 35%` of `CH_(4), 50%` of `H_(2)` and `5%` of non-combustible gas. The air contains `20.8%` (by volume) of oxygen. |
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Answer» Correct Answer - `5m^(3)` `2H_(2)O+O_(2)to2H_(2)O` `CH_(4)+2O_(2)toCO_(2)+2H_(2)O` `2CO+O_(2)to2CO_(2)` `C_(2)H_(2)+5/2 O_(2)to 2CO_(2)+H_(2)O` Required oxygen of the combustioi of given gas `=1.04m^(3)` Hence, volume of air required `=(1.04xx100)/20.8=5m^(3)` |
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