2.70 mol C2H4 x (3 mol O2/1 mol C2H4) = 8.1 mol O2 required. Note how the conversion factor I used converts mol C2H4 to moles O2 by dividing out (canceling C2H4 units and leaving unit O2). Do you have 8.1 moles O2? No. Therefore, all of the C2H4 can't be used because O2 is the limiting reagent. That is, there is not enough oxygen to burn all of the C2H4. You can check this (or suppose we had chosen the oxygen to start).
6.3 mol O2 x (1 mol C2H4/3 moles O2) = 2.1 mol C2H4 needed to react with all of the oxygen. Do you have that much C2H4? Yes, we have 2.70 mols C2H4, which is enough to react with all of the oxygen and have some un-reacted C2H4 (2.70-2.10 = 0.60 mol) remain.
The equation for the complete combustion of ethene (C2H4) is
C2H4(g) + 3 O2(g) ==> 2CO2(g) + 2H2O(g)
If 2.70 mol C2H4 is reacted with 6.30 mole O2, identify the limiting reagent.
show all work.
3 answers
2H2+o2>2h2o
heck it up
1 answer