Question

\( 80 cm ^{3} \) of methane is mixed with \( 200 cm ^{3} \) of pure oxygen at the same temperature and pressure. The mixture is then ignited. Calculate the composition of the resulting mixture if it is cooled to initial room temperature and pressure. \[ \begin{array}{c} CH _{4}+2 O _{2} \longrightarrow CO _{2}+2 H _{2} O \\ \left(\text { Ans : } CO _{2}=80 cc , O _{2}=40 cc , H _{2} O =\text { negligible }\right) \end{array} \]

Answer 1

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

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

1 volume of CH₄ react with 2 volume of pure oxygen (O₂) to form 1 volume CO₂.

(Volume of H₂O(l) is negligible to compare volume of gaseous molecules)

∴ Here, CH₄ is limiting reagent.

∴ Volume of pure O₂ unreacted = (200 - 160)cm³ = 40 cm³

∴ Volume of CO₂ formed = 80 cm³

Hence, mixture contain 40 cm³ O₂ and 80 cm³ CO₂.