# $80 cm ^{3}$ of methane is mixed with $200 cm ^{3}$ of pure oxygen at the same temperature and pressure.

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$80 cm ^{3}$ 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. $\begin{array}{c} CH _{4}+2 O _{2} \longrightarrow CO _{2}+2 H _{2} O \\ \left(\text { Ans : } CO _{2}=80 cc , O _{2}=40 cc , H _{2} O =\text { negligible }\right) \end{array}$

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

∵ 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.