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Author: Subject: stoichemetry question
CHRIS25
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[*] posted on 12-3-2014 at 01:52
stoichemetry question


I have been away from chemistry for almost 18 months and am embarrassed to say that I have forgotten some basics. My notes have confused me on this one thing:

Which is the correct procedure?

Cu+2HCl=CuCl2+H2 (for etching)

63g copper + 1/11 (M concentration of HCl)) =90mLs
therefore approx 180 mLs of 11M conc HCl needed.

OR

63g copper needs 72g/mol/1.2 (density of approx 36%HCl)
therefore 60 mLs of 11M soln HCl needed.

I am embarrassed but after 18 months I am trying to revise through all my notes and have looked at my previous posts but could not find the answer.

Thankyou and sorry for the bother here.




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[*] posted on 12-3-2014 at 02:36


I don't think the equation is right. You need an oxidant. hence the equation is Cu(s) + H2O2(aq) + 2HCL(aq) ----> CuCl2 (aq) + 2H2O(l).

I won't do it for you but remember:
n=m/M (solids) and n=cV (for liquids)
Where n= no.of moles. m=mass of sample. M=Molar mass
n=no.of moles. c= concentration. V=volume.
So:
M(Cu)= 63.546 g
M(HCL)= 36.46094 g
M(H2O2)=34.0147 g


From The equation: n(HCL)/n(Cu)= 2:1 ratio

Hence per one mole of copper we need two moles of HCL.
If you have the % concentration you don't even need the density.
I have to go, but i hope this helps; i'm sure you can manage.

[Edited on 12-3-2014 by HeYBrO]




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CHRIS25
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[*] posted on 12-3-2014 at 02:48


Yes I know, I use air, or H2O2. I solved it, 72g HCl are needed This is 2moles (11M is my concentration). 2/11 = 0,189 L = 190mLs of 11M HCl. Thanks for the reminder, I have no idea what I am doing dividing g/mol by the density.



‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)

The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by precision and law. (me)
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[*] posted on 12-3-2014 at 05:54


A great thing I learned in physics that helps me out all the time is dimensional analysis. Just look at the units of the quantities you know, and the units of the quantity you want. Then arrange things so that the correct final units fall out. You're usually only off by a constant, which you can easily figure out on a multiple choice test :)

Dividing molecular weight by density gives you (g/mol) * (mL/g) = mL/mol , which is a good clue that something is amiss!
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CHRIS25
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[*] posted on 12-3-2014 at 10:00


Is this a better way: What volume of 36% HCl needed to contain 72g?

100mLs * 72g / 36% = 200mLs of 36%HCl contains 72g?

I am so rusty, desperately trying to re-learn everything.




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)

The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by precision and law. (me)
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[*] posted on 12-3-2014 at 10:39


36% usually means 36% by weight.
100 mL would have 36 g of HCl, so yes, 200 mL is going to contain double.




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[*] posted on 12-3-2014 at 12:14


Thankyou vmelkon



‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)

The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by precision and law. (me)
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