Chem(Ideal Gas Law)

(i)
At 130°C :
CH4(g) + 2O2(g) → CO2(g)+ 2H2O(g)

When 80 cm³ of CH4 reacts,
volume of O2 needed = 80 x 2 = 160 cm³
Hence, CH4 completely reacts (limiting reactant), and O2is in excess.

Volume of O2 left = 180 - 160 = 20 cm³
Volume of CO2 formed = 80 cm³
Volume of H2O(g) formed = 80 x 2 = 160 cm³

Volume of residual gas = 20 + 80 + 160 = 260cm³


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(ii)
At 130°C :
CH4(g) + 2O2(g) → CO2(g)+ 2H2O(l)
At 25°C, the water formed is in liquid state, and thus its volume becomesnegligible.

When 80 cm³ of CH4 reacts,
volume of O2 needed = 80 x 2 = 160 cm³
Hence, CH4 completely reacts (limiting reactant), and O2is in excess.

Volume of O2 left = 180 - 160 = 20 cm³
Volume of CO2 formed = 80 cm³

Volume of residual gas = 20 + 80 = 100cm³

(i)
by Avogadro's Law, volume of gas is directly proportional to the gas volume under constant temperature and pressure.
chemical equation:
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g) (note: water exsist in vapor phase under 1 atm & 130°C)
so 80cm³ CH4(g) reacts with 160cm³(g) of O2 (20cm³ in excess)
gives 80cm³ of CO2(g) and 160cm³ H2O(g) (240cm³ gaseous product)
so, including the excess 20cm³ of O2 , we have 260cm³ residual gas.

(ii)
similar to part (i), but note that, under 25°C, water condenses.
so the chemical equation becomes:
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)
so 80cm³ CH4(g) reacts with 160cm³(g) of O2 (20cm³ in excess)
gives 80cm³ of CO2(g)
including the excess 20cm³ of O2, total volume of residual gas = 100cm³