"For 4HCl + O2 <-> 2Cl2 + 2H2O :
If in a container of 1.0 L we start with 2 moles of O2 and 4 moles of HCl, the molar fraction of O2 is:
a) x/5 - x/2
b) 1-x / 6+x
c) 5x / 5-x
d) 1-x / 5-x
e) 2x / 6-x
"
I tried solving this, but I wasn't sure if all 2 moles of O2 were used or not, because of the proportion established in the equation. I tried doing it with 2 moles of O2 (got 2-x / 6-2x) and 1 mol of O2 (got 1-x / 5-2x) yet neither of the two results I got were in the choices. I'm not sure either if I have to count the moles of the products or not?
At first I thought the moles of the products didn't count, but when I tried solving it with them and using 1 mol of O2, I got (1-x / 5-x) which is one of the choices. Not convinced it's the correct one though.
They want O2 after equilibrium is reached. The problem doesn't give me the equilibrium constant.
And I suppose x would be the unknown value you put when you're writing the ICE table? for example there are 2 moles of O2 at the beginning, x would be the change, and the value in equilibrium would be 2-x.
This is what I've done so far
4HCl + O2 <-> 2Cl2 + 2H2O
4 2
-4x -x 2x 2x
4-4x 2-x 2x 2x
And to find the molar fraction of O2 it'd be something like 2-x / 4-4x+2-x+2x+2x (moles of O2/total of moles) = 2-x / 6-x
Or in the case only 1 mol of O2 is used then i suppose it'd be something like
4HCl + O2 <-> 2Cl2 + 2H2O
4 1
-4x -x 2x 2x
4-4x 1-x 2x 2x
1-x / 4-4x+1-x+2x+2x
= 1-x / 5+x
As you can see, none of the fractions I got are in the choices
2-x/6-x is correct.
I suspect the answer is d and a typo was made in which the negative sign was omitted between he 2 and x.
I went back to look at my procedure and I noticed I made an error in the second one, I actually got (1-x / 5-x) , which is one of the choices.
Could that be the answer? Or do we have to take into account all moles of O2 despite the proportion given in the equation, thus the correct answer would have to be the first one (2-x/6-x)?
I'm not sure if stoichiometry works the same here in equilibrium and if we have to take into account the limiting reagent.
If this is an equilibrium process AND the reaction doesn't go to completion, the correct answer is 2-x/6-x. I assume, as you have assumed, that the x is like the ICE method. If that is the case, you start with 2 and subtract the x and what's left is what is left over. If you want the mole fraction of O2 AFTER equilibrium, then 2-x will be the mols O2 after equilibrium is established. Just as 4-4x is mols left of HCl, then 2-x is mols O2 left and 2x is mols Cl2 formed. XO2 = mols O2/total mols.
So all 2 moles of O2 are used for the molar fraction formula even though only one of its moles can react to the 4 moles of HCl because '-x' is being added to it?
Actually, you don't know how many mols O2 are used nor do you know how many mols HCL are used. Since the reaction is at equilibrium, and we don't know the equilibrium point since there is no Keq, all we know is that we started with 2 mols O2, we have used x amount, which leaves us with 2-x mols O2 remaining. Likewise, we don't know exactly how much HCl is used but we started with 4 mols, we have used 4x so we are left with 4-4x remaining. We don't know how much Cl2 and H2O are formed, either, but the best we can do is to say they are 2x. All of this leaves a funny way of expression XO2. If the reaction had gone to completion (ig it has a large K but the problem gives us no clue what that is), we would know there was 1 mol O2 left, no HCl, 2 mols CL2 and 2 mols H2O formed and XO2 would be 1/5. Hope that clears things up.
I see, thank you so much for the help! I didn't expect anyone to answer and less reply continuously haha. I'm very grateful! Thank you once again and have a great day!
You're welcome. We do our best to help. Jiskha is a great site, I think. Remember there are experts here in SS, history, English, math, physics.
To determine the molar fraction of O2 in the given chemical equation and starting with 2 moles of O2 and 4 moles of HCl in a 1.0 L container, we need to consider the balanced equation and the stoichiometry.
The balanced equation is: 4HCl + O2 -> 2Cl2 + 2H2O
From the equation, we can see that it requires 1 mole of O2 to produce 2 moles of Cl2 and 2 moles of H2O. Therefore, for the reaction to proceed completely, we would need 4 moles of O2.
However, we have only 2 moles of O2. This means that O2 is present in excess. In this case, the limiting reactant is HCl, and the amount of Cl2 and H2O formed will be determined by the amount of HCl present.
Now, let's calculate the molar fraction of O2:
Molar fraction = moles of O2 / total moles of all components
Total moles of all components = moles of O2 + moles of HCl
Moles of O2 = 2
Moles of HCl = 4
Total moles of all components = 2 + 4 = 6
Molar fraction of O2 = 2 / 6 = 1/3
Therefore, the molar fraction of O2 is 1/3.
Now let's check which option matches this result from the given choices:
a) x/5 - x/2
b) 1-x / 6+x
c) 5x / 5-x
d) 1-x / 5-x
e) 2x / 6-x
The correct answer is option c) 5x / 5-x.
It's important to note that in stoichiometry problems like this, we consider the balanced equation and the stoichiometry to determine the limiting reactant and the amount of products formed. Excess reactants are not fully consumed, but they may affect the final result in terms of limiting reactants and reactant ratios.
I can't help. This is a TERRIBLE problem.
1. Do you want XO2 before reaction or after reaction.
3. If after reaction, does the rxn go to completion.
4. If it doesn't go to completion what is the equilibrium constant.
5. What the H..L is x?
If after reaction you count products, too, if the rxn goes to completion.
Not all of the O2 is used.