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)+20_(2)(g)to CO_(z)(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 gCO_(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 22g `CO_(2)(g)`