Suppose that 3.2 L of ethene gas (C2 H4 ) at
1 atm and 298 K is mixed with 8.6 L of oxy-
gen gas at the same pressure and temperature
and burned to form carbon dioxide gas and
liquid water. Ignoring the volume of water
forward, ﬁnd the ﬁnal volume of the reaction
mixture (including products and excess reac-
tant) at 1 atm and 298 K if the reaction goes
Answer in units of L
For #1, P1V1 = P2V2
C2H4 + 3O2 ==> 2CO2 +2H2O
3.2L C2H4 will form 2*3.2 = 6.4L CO2 if it has all of the O2 needed.
8.6L O2 will form 8.6 x (2 mols CO2/3 mols O2) = 5.73 mols CO2 if it has all of the C2H4 needed.
Therefore, O2 is the limiting reagent and you will have 5.73 L O2 formed. We ignore the volume of H2O.l
How much C2H4 is used? That is 8.6L O2 x (1 mol C2H4/3 mol O2) = 2.87 mols used. We had 3.2 initiall; there must be 3.2-2.87 = 0.33L left over.
Total volume = 0.33L C2H4 + 0 oxygen + 5.73L CO2 + ignored H2O = ?
thank you very much! i'll look into redoing the question and seeing if we get the same answer. Thank you for your time
for the first question, how do you find P2. because i understand that the kpa increases by 100 every 10.2 meters but i don't know how to fimd it for 35.78 meters ???
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