Quick question on Solution Stoichiometry

[betterfuture]

I was doing a problem on limiting reagent when dealing with solutions and was wondering when one finds the limiting reagent, why it is incorrect to add the total volume together to find the moles of the reactants?

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Example: A 25 mL sample of 1.20M KCl solution is mixed with 15 mL of 0.900M Ba(NO3)2 solution. BaCl2 is collected, dried and found to weigh 2.45 g. Find the limiting reagent, theoretical yield, and percent yield.

2KCl + 1Ba(NO3)2 --> 1BaCl2 + 2KNO3

Method:

Figuring out moles of KCl (my way)

(1.20M) * (.025L+.015L) = .048 moles KCl * (1 BaCl2/2 KCl) = .024 BaCl2 (incorrect)


Book:
(1.20M) * (.025L) = .03 moles KCl * (1 BaCl2/2 KCl) = .015 BaCl2 (correct)
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Shouldn't you add the total volume together and then find the moles or is this method incorrect? Can someone give a reason to why you can't do it the way I did? Thank you!

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[theonlytycrane]

The KCl only comes from the 25 mL of 1.20M solution, and Ba(NO3)2 only comes from the 15 mL of .900 Ba(NO3)2 solution. In the end you have one total volume, but to figure out how many mol of each reactant you start with you only want to use the volume from which they came from.

 

[betterfuture]

So if you wanted to find the molarity of KCl in the solution after Ba(NO3)2 was mixed, then you would total the volumes, correct?

Because molarity depends on total volume while moles does not depend on volume and on the molar mass of the molecule only?

 

[theonlytycrane]

Yes, for molarity.

mol just depends on using the relationship Molarity = mol / V with each individual reactant.

 

[betterfuture]

But isn't moles a fixed number. Like, if you increase the volume or molarity, the moles stay the same?

 

[theonlytycrane]

Think of the KCl example. The number of mol is determined from the 25 mL of 1.20M solution. If I add more of it (more volume) or use a solution with higher concentration (say 1.50M), than the moles of KCl will be increased.