A mixture of CH4 and C2H2 occupied a certain volume at a total pressure of 63 mm.The sample was burnt to CO2 and H2O and the CO2 alone was collected and it's pressure was found to be 69 mm in the same volume and at the same temperature as the original mixture.What fraction of the mixture was Methane ??

Dear ,
1 answer ? Chemistry

Best Answer
from this unbalanced equation
CH4 + C2H2 + O2 ? CO2 + H2O
if we let moles of CH4 be x and moles of C2H2 be y
then moles of CO2 will be (x+2y) to account for all the carbons
PV = nRT but V, R an T are all constants in this equation
P = n Z where Z is the collection of constants
nZ on the left = 63Z = x+y
nZ on the right = 96Z = x+2y
subtracting the first equation from the second
33Z = y so x=30Z because 63Z is the total
CH4/100% = part/whole = 30Z / 63Z = 0.476
CH4 = 47.6% of the original sample.

i hope this help you...
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Co2 alone was collected and its pressure is 96 mm ,instead of 69mm
  • -1
Dear, CH4 +2O2 ----CO2 +2H20 C2H2 +5/2O2 ----2CO2 +H20 let the partial pressure of CH4 be p thus the partial pressure of C2H2 is 63-p thus following the question total pressure of CO2 after combustion=69 mm therefore p +2(63-p) =69 p=57 mm partial pressure of C2H2=6mm percentage of CH4=57/63*100 =90.41% i hope this help you i phope
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