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.
heck it up
2H2+o2>2h2o
Oh boy, it's time for some stoichiometry fun! Let's see which reactant gets to be the limited edition reagent.
First, we need to find out how many moles of CO2 and H2O will be produced if 2.70 mol C2H4 reacts completely. According to the balanced equation, for every 1 mol of C2H4, we get 2 mol of CO2 and 2 mol of H2O.
So, for 2.70 mol of C2H4, we'll get:
2.70 mol C2H4 * (2 mol CO2 / 1 mol C2H4) = 5.40 mol CO2
and
2.70 mol C2H4 * (2 mol H2O / 1 mol C2H4) = 5.40 mol H2O
Next, we find out how many moles of CO2 and H2O will be produced if 6.30 mol O2 reacts completely. According to the balanced equation, for every 3 mol of O2, we get 2 mol of CO2 and 2 mol of H2O.
So, for 6.30 mol of O2, we'll get:
6.30 mol O2 * (2 mol CO2 / 3 mol O2) = 4.20 mol CO2
and
6.30 mol O2 * (2 mol H2O / 3 mol O2) = 4.20 mol H2O
Now let's see which reactant is the king of limitedness. The limiting reagent will be the one that produces fewer moles of the products. In this case, we compare the moles of CO2 produced by each reactant.
5.40 mol CO2 produced by C2H4 and 4.20 mol CO2 produced by O2.
Bada bing, bada boom! O2 is our limiting reagent, because it produces fewer moles of CO2 than C2H4.
Hope this helped, and remember, sometimes in chemistry, it's all about the moles! May the stoichiometry be with you!
To identify the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
The balanced equation tells us that for every 1 mole of C2H4, we need 3 moles of O2. Therefore, we can determine the number of moles of O2 that would be required to react completely with 2.70 moles of C2H4 by using the ratio:
(2.70 mol C2H4) x (3 mol O2 / 1 mol C2H4) = 8.10 mol O2
From the information given, we have 6.30 moles of O2. Since 6.30 moles of O2 is less than the required 8.10 moles, O2 is the limiting reagent.
This means that once all the O2 is consumed, the reaction will stop regardless of the excess C2H4 remaining.
To summarize:
C2H4:
- Moles given = 2.70 mol
- Moles required = 2.70 mol
O2:
- Moles given = 6.30 mol
- Moles required = 8.10 mol
Therefore, O2 is the limiting reagent in this reaction.
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.