Stoichiometry problems!
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Stoichiometry problems!

[From: ] [author: ] [Date: 12-02-03] [Hit: ]
10 moles of O2.Note the mole ratio between O2 and KClO3.It is 3 to 2.OR for every mole of O2 produced you will only need 2/3 of that many moles of KClO3So 10 x 2/3 will give you the answer.2.15 g of KClO3 .......
I don't know how to do this stuff at all. I've linked my review and If someone could just answer one of the questions and explain it to me! Please

http://tinypic.com/view.php?pic=125n7ed&…

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2 KClO3 -----------> 2 KCl + 3 O2

1. 10 moles of O2. Note the mole ratio between O2 and KClO3. It is 3 to 2. OR for every mole of O2 produced you will only need 2/3 of that many moles of KClO3 So 10 x 2/3 will give you the answer.

2. 15 g of KClO3 . Change that to moles. moles = 15 g / 122.55 g/mol = 0.1224 moles

Mole ratio between KClO3 and KCl is 2 to 2 or 1 to 1 so you will produce 0.1224 moles of KCl

3. 5 moles of KClO3 The mole ration says 3 to 2 or 3/2 so 5 x 3/2 = 7.5 moles of O2 Assuming STP conditions then the volume will be 7.5 moles x 22.4 L/moles = 168 L

4. 2.5 g of KClO3 . Change that to moles 2.5 g / 122.55 g/mol = 0.0204 moles

Mole ration is 1 to 1 so 0.0204 moles of KCl. Not just change that back into a mass

mass = moles x molar mass. I'll leave you do the last calculation.
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