How many moles of methane are required to produce `22 g CO_(2)(g)` after combustion ?

Explanation

According to the chemical equation,
`CH_(4(g)) +2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((g))`
44 g `CO_(2(g))` is obtained from 16 g `CH_(4(g))`
( `:.` 1 mol `CO_(2(g))` is obtained from 1 mol of `CH_(4(g))` )
Mole of `CO_(2(g)) = 22 g CO_(2(g)) xx ("1 mol" CO_(2(g)))/(44 g CO_(2(g)))`
`= 0.5 "mol" CO_(2(g))`
Hence, 0.5 mol `CO_(2(g))` would be obtained from 0.5 mol `CH_(4(g))` or 0.5 mol of `CH_(4(g))` would be required to produce 22 g `CO_(2(g))`.