\(\underset{initially}\,\,\,\;\underset{80cm^3}{CH_4(g) }+ \underset{200cm^3}{2O_2(g)} \longrightarrow CO_2(g) + 2H_2 O(l)\)

∵ Temperature and pressure as constant, from the above chemical equation we can conclude that -

1 volume of CH_{4} react with 2 volume of pure oxygen (O_{2}) to form 1 volume CO_{2}.

(Volume of H_{2}O(l) is negligible to compare volume of gaseous molecules)

∴ Here, CH_{4} is limiting reagent.

∴ Volume of pure O_{2} unreacted = (200 - 160)cm^{3} = 40 cm^{3}

∴ Volume of CO_{2} formed = 80 cm^{3}

**Hence, mixture contain 40 cm**^{3} O_{2} and 80 cm^{3} CO_{2}.