If 29.0 L of methane CH4 undergoes complete combustion at 0.961atm and 20°C ,Howmany liters of each product are formed?
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Expert's answer
2018-05-09T07:16:52-0400
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
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