anonymous
  • anonymous
A 0.1647g of hydrocarbon sample is going through combustion process. It produces 0.5694g of CO2 and 0.0845g of H2O. Determine the percentage of these elements in this hydrocarbon.
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
CxHy + 3/2 O2------> xCO2 + Y/2H20 take out mole of CO2 as u are provided mass and molecularweight ok
anonymous
  • anonymous
hydrocarbon is CH4 i get that but now how to cal % ?
anonymous
  • anonymous
any idea?

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More answers

anonymous
  • anonymous
hey i got C3H6
anonymous
  • anonymous
no how can be this possible i got x=1 and y=4 by that i cal hydrocarbon results CH4
anonymous
  • anonymous
u show me ur calculus and i ll show mine ok
anonymous
  • anonymous
K
anonymous
  • anonymous
CxHy + 3/2 O2------> xCO2 + Y/2H20 take out mole of CO2 as u are provided mass and molecular weight x=0.5694/44 x= approx 1 let y mole of H2O y=0.0845/18 y=4 mole og h20 =y/2=2
anonymous
  • anonymous
ur turn
anonymous
  • anonymous
I WAS DOING Y/2= 0.004 => Y= 0.008 LOL
anonymous
  • anonymous
wat funny :P anyways nw how to find percentage :?
anonymous
  • anonymous
I WILL NOW SOLVE % COMPOSITION WAIT!!
anonymous
  • anonymous
ok :) can i go now ? i have to do revision of hydrocarbon
anonymous
  • anonymous
H ---> 75% C-25% AM I RIGHT??
anonymous
  • anonymous
how u cal % ? frankly saying i forgot all things :( plz show me ur work
anonymous
  • anonymous
sry sry H-25% C-75%
anonymous
  • anonymous
for H , 4/16 x 100= 25 % \ for C 12/16 x 100 = 75%
anonymous
  • anonymous
yup :)
anonymous
  • anonymous
hi5 we solve it :)
anonymous
  • anonymous
i alwz do silly mistake sry
anonymous
  • anonymous
even i too not a big prob its d symbool dat we are still learners:)
anonymous
  • anonymous
i have a different aprocah for this problem , shall i tell u
anonymous
  • anonymous
sure :)
anonymous
  • anonymous
1 mole of CO2----> 44 g => 1 g of CO2---> 1/44 mole => 0.5694 g of CO2--> 1/44 x 0.5694 =.01294
anonymous
  • anonymous
i knw this method :)
anonymous
  • anonymous
same thing for H20 == 0.004694 mole
anonymous
  • anonymous
:-) great
anonymous
  • anonymous
can i go now? plz dont mind but its urgnt for me to revise it as i dont knw the abc of alkene ... sorry
anonymous
  • anonymous
yea sure.... bye tnx for the help

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