Question

If 29.0 L of methane CH4 undergoes complete combustion at 0.961atm and 20°C ,Howmany liters of each product are formed?

Answer 1

First convert the given volume of CH4 into S.T.P
P1=0.961atm. P2=1 atm.
V1=29.0 L V2=?
T1=(273+20) K = 293K T2=273K
Now by combining Boyle law and Charles law
P1*V1/T1 = P2*V2/T2
On putting values and solving V2=25.9667 L
Now the reaction of combustion of methane:
CH4 + 2O2 = CO2 + 2H2O
Now, 1mole CH4 gives 1 mole CO2
also, 22.4L CH4 gives 22.4 L CO2
Therefore 25.9667 L CH4 gives 25.9667L=26L CO2
Similarly,
1 mole CH4 gives 2 moles H2O
also 22.4 L CH4 gives 2*22.4 L H2O
Therefore, 25.9667 L CH4 gives 2*25.9667 L of H2O=51.9334 L =52 L H2O