Question

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.

Answer 1

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)`