ZoSo357
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Registered: 30-8-2005
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Gravimetric stoichiometry
I didn't see anything in the first ten pages of this section of the forum about stoichiometry, so perhaps this could be a great place for others with
trouble doing stoich to post their questions and help each other.
Anyways, here goes.
I was doing some stoichiometry on my own, and my goal at the time was to balance the oxygen of ETN with Al. I realized i wasn't sure how to do this,
and was using a HE calculator, and figured i'd try on my own. I started with a formula i knew, I knew that the standard flash comp of KClO4/Al was
70/30 so I used this for a reference.
I started by making myself a problem like you tend to get in school.
"How many grams of KClO4 would be needed to react with 30 grams of aluminum"
Naturally the answer should be about 70.
My workings were as follows: (By the way, All of this isn't for sure to me, so I'm sorry for any stupid mistakes)
Key:
m- mass and/or weight
n- moles
M- molar mass
KClO4 + 2Al --> KCl + Al2O3 + 1-O
----------- ^
-----------30g
Molar mass of Al - 26.98
Molar mass KClO4 - 138.547
K 1x-39.098 = 39.098
Cl 1x-35.453 = 35.453
O 4x-15.999 = 63.996
= 138.547
mols(Al) = mass/molar mass = 30g / 26.98 = 1.111mols Al
(I then calculated mols of KClO4)
1.111 mols of Al X 1-KClO4/2Al = 0.5555 mols KClO4
next step was finding my mass of KClO4:
m=n X M = 0.5555 X 138.547 = 77g
______________________________________________
So by my equation, I get a 77 to 30 ratio. I know this isn't the commonly used ratio (which i believe is O balanced). So does this mean that the
decimals i'm using aren't high enough, or am I doing something wrong, and just fluking into having a near-70 mass?
Basically I want to be able to balance to oxygen in ETN with aluminum, but i want to be able to do it on my own and balance other things
independantly, so the first step I guess is asking others. Thank you in advance for the help!
EDIT: By the way, in my equation:
KClO4 + 2Al --> KCl + Al2O3 + 1-O
----------- ^
-----------30g
Those hyphens (- ) are there because otherwise i couldn't place the 30g under the Al.
[Edited on 10-12-2006 by ZoSo357]
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unionised
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Registered: 1-11-2003
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There's an "O" left over in your equation. I'm pretty sure that the whole point of oxygen balance is to avoid that.
The first thing to do is balance the equation.
3 KClO4 + 8 Al ----> 3 KCl +4Al2O3
So
3*138.5 + 8*27 ---->3*74.5 + 4 *102
415.5 + 216 ----> 223.5 +408 (always worth working out the masses of the products to check it balances)
Each 216g of Al reacts with 415.5g KClO4 so each gram of Al takes 415.5/216 ie 1.92 g of KClO4
30 g would need 57.7g
That seems a bit low, I may have got something wrong or the "usual mixture" might not be stoichiometric
There's another way to do it (though you still need the equation)
You have 30g of Al which is 1.11 moles
From that equation each mole of Al requires 3/8 moles of KClO4
1.11*3/8*138.5= 57.7g
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