I'm struggling with stoichiometry in regards to the way moles relates to % solution.
For example, Cu + 2HCl -> CuCl2 + H2
So I need one mole of Copper, so i'll use one molar, or 63.5g
Now the HCl I have is normal hardware grade ~31.45% HCl. For the above formula, would I try and make a 2 molar solution of this 31.45% HCl? Or would I
add an excess of 68.55% (100-31.45) so it would be like the HCl was 100%? I hope this question makes sense, it's pretty hard for me to put it in the
right words.MrHomeScientist - 28-12-2018 at 07:24
It's easier to work in molarity than percent. I recommend titrating your hardware store acid to determine molarity and go from there. For example, my
hardware store HCl came out to be 9.5M, or 9.5 mol/L.
So, if you want to start with 1 mole of copper, you're right that you would need 63.5g. To fully react this, you'd need:
1 mol Cu * (2 mol HCl / 1 mol Cu ) * (1 L / 9.5 mol ) * (1000 mL / 1 L) = 210.5 mL HCl
It's usually a good idea to use a bit excess of acid to ensure all the copper is dissolved. You'll probably be left with some dust that won't
dissolve, from impurities in the copper. Just filter that off and allow the water to evaporate to yield your copper chloride crystals, assuming that's
what your're after.happyfooddance - 28-12-2018 at 07:28
Maybe a bigger problem is that copper and HCl don't react
But as for your stoichiometry, 1 mol of HCl is about 36.5 grams. 2 mol is 73 grams. If your acid is about 32% then that is about 225 grams of acid.
You have to do a little math.r0749547 - 29-12-2018 at 11:54
He could use H2O2 to make it react
or bubble air through it.
[Edited on 29-12-2018 by r0749547]thehexlibrary - 29-12-2018 at 18:33
Thanks guys so much for the help so far!
In addition to titration, one could make calculations based off of the %v/v or %w/v? Like if we have 9.5M HCl, we could do the math:
1L * (9.5 mol / 1 L)*(36.5g HCl / 1 mol) = 346.75g HCl per L. So a 9.5M would be 34.6% HCl? Is this correct?
Also, MrHomeScientist love your vids man!Tsjerk - 30-12-2018 at 01:05
In addition to titration, one could make calculations based off of the %v/v or %w/v? Like if we have 9.5M HCl, we could do the math:
1L * (9.5 mol / 1 L)*(36.5g HCl / 1 mol) = 346.75g HCl per L. So a 9.5M would be 34.6% HCl? Is this correct?
34.6% is correct for weight/volume, but usually the percentage concentration for HCl means weight/weight, so you have to factor in the density of 9.5M
HCl (1.15 kg/L).
This would mean: 346.75g of HCl per liter = 346.75g per 1.15 kg = 30.2%
This makes sense as 37% HCl is about 12M, and 30% HCl is a common concentration found in hardware stores.
I guess most people would agree molarity is easier to work with, as confusion easily sets in when working with w/w, w/v and or v/vhappyfooddance - 30-12-2018 at 11:59
I disagree about molarity being more convenient than w/w.
Molarity changes when solutions are mixed, and their densities fluctuate (often non-linearly).
Keeping track of w/w percentages is far more accurate and less messy record-keeping-wise, and when working with mixtures is the only sensible way to
go.
But it also goes hand in hand with the type of chemistry I do most (food and flavors) so that might also be it. Not saying what is best for anyone
else, just myself.
Edit: I guess it also depends on what tools you use to measure. I practically live next to the analytical balance. I do use a fair bit of volumetric
glassware, but it is almost always placed on a balance at some point.
