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Topic: mols  (Read 7072 times)

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« on: May 27, 2004, 10:23:27 PM »
I have a test on mols, heres a practice problem:

2KClO3 -----> 2KCl + 3O2
25 grams of potassium chloride react to produce how many grams of oxygen?

I said 32.43g but dont know if its right. please help. god please. and tell me how to do it if i did it wrong. please god.

i converted the kcl on the product side 25 grams to .337 mols and multiplied that by 32g (3O2) to get 10 .81 grams and to get it in ratio with the others i just multiplied it by 3 (mols of oxygen in product) to get 32g
« Last Edit: May 27, 2004, 10:40:19 PM by berryblue »

Offline Mitch

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« Reply #1 on: May 27, 2004, 11:31:12 PM »
Are you sure the problem wasn't "25 grams of potassium perchlorate react to produce how many grams of oxygen?
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« Reply #2 on: May 28, 2004, 03:42:40 PM »
I am absolutely sure that you would not start with 25 grams of substance, which then undergoes decomposition to make 32 grams of something else.  There are physical laws that look suspiciously like "conservation of matter" that interfere with this happening.

Offline hmx9123

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« Reply #3 on: May 28, 2004, 08:00:53 PM »
KClO3 is potassium chlorate.  (Potassium perchlorate is KClO4)  

Anyway, to get to your question:

25g of potassium chlorate is how many moles?  You find out by dividing 25g by the formula weight of potassium chlorate.  Once you have this number of mols, then you use the stoichiometry of the chemical reaction to give you the number of mols of your product.  In your case, your reaction uses 2 mols of potassium chlorate to produce 3 mols of oxygen gas, so there is a 2:3 ratio of mols of potassium perchlorate used: mols oxygen gas generated.  Once you calculate the number of mols of oxygen gas you have, then you use the formula weight of oxygen gas (32 g/mol) to give you the number of grams of oxygen that you generated.

I think that you might be misreading the question a little; potassium chloride won't decompose to give oxygen--only potassium chlorate will do that (in your equation).

Hope this helps.

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