# Chemistry

If I have 50.0 mL of HA at a concentration of 1.20M and 50.0 mL of NH4OH at a concentration of 0.92M, how many moles of the limiting reactant are there?
I guess I should start with a balanced equation, but I have no idea where to start..

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1. mols HA = M x L = 1.20 x 0.050 = 0.06
mols NH4OH = M x L = 0.92 x 0.05 = 0.046
..........HA + NH4OH ==> NH4A + H2O
I.......0.06..0.046.......0.......0
C.......
E.......
These two materials react in a 1:1 ratio so which one looks as if it is the limiting regent; i.e., which will be used up first? Which will have some that didn't react?

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2. Looks like 0.046 mol NH4OH.
So, I just multiply the moles and liters and then compare the two values? The balanced equation isn't really necessary?

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3. Yes, the balanced equation is VERY necessary. Otherwise you don't know that they are reacting in a 1:1 ratio. And you want to be careful when comparing the numbers. For example, suppose we had the same problem and numbers EXCEPT we used H2A and NH4OH.
.......H2A + 2NH4OH ==> (NH4)2A + 2H2O
I.....0.06...0.046........0........0
C.....
E.....
The 1:1 ratio makes it an easier problem. In this problem you must actually calculate mols used for each; i.e., how much NH4OH would it take to react with ALL of the H2A.
That's 0.06 mols H2A x (2 mols NH4OH/1 mol H2A) = 0.06 x 2/1 = 0.12. Do you have that much NH4OH? No, therefore, NH4OH must be the limiting reagent (just as in the previous problem). You can check that by asking how much H2A is necessary to react with ALL of NH4OH? That's 0.046 mols NH4OH x (1 mol H2A/2 mols NH4OH) = 0.046 x 1/2 =0.023 mols H2A. Do you have that much H2A? Yes, you have more than enough; again, NH4OH is the limiting reagent. Even though the answer for the limiting reagent came out the same, you can see that this example does not allow you to compare the numbers as is; that is, they must be "corrected" for those coefficients if they are not 1:1.

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