Asked by rekha
132gram o2 is required for a combustion of 28 liter mixture of ethane and ethene (at NTP) than find out mole fraction of ethane and ethene in the mixture?
Answers
Answered by
DrBob222
I don't believe this problem is workable as is. Here is why.
If you use NTP, then mols in 28L is 28/24.05 = 1.164
Suppose ALL 1.164 mols are present as C2H6, then g O2 required is
1.164 x (7 mols O2/2 mols C2H6) x (32 g O2/1 mol O2) = about 130 grams O2 which comes close but doesn't use all of the oxygen. Equation is
2C2H6 + 7O2 ==> 4CO2 + 6H2O
Therefore, you need to use more oxygen but you no more moles available to do so. I think that 132 g must be somewhere between 112 and <130.4 g.
If you use NTP, then mols in 28L is 28/24.05 = 1.164
Suppose ALL 1.164 mols are present as C2H6, then g O2 required is
1.164 x (7 mols O2/2 mols C2H6) x (32 g O2/1 mol O2) = about 130 grams O2 which comes close but doesn't use all of the oxygen. Equation is
2C2H6 + 7O2 ==> 4CO2 + 6H2O
Therefore, you need to use more oxygen but you no more moles available to do so. I think that 132 g must be somewhere between 112 and <130.4 g.
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