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Topic: How many moles of Cl2...?  (Read 10541 times)

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Offline horsegurl221

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How many moles of Cl2...?
« on: July 21, 2011, 02:23:52 PM »
I'm trying to figure out how many moles of Cl2 are in the product side of this equation:
2HMnO4+ 14HCl --> 2MnCl2+ 5Cl2+ 8H2O
I know that the HCl is the limiting reactant and 2MnCl2 is 0.5714285714 moles, but how do I figure out how many moles are in the Cl2 part of the 2MnCl2???
« Last Edit: July 21, 2011, 02:41:24 PM by horsegurl221 »

Offline DevaDevil

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Re: How many moles of Cl2...?
« Reply #1 on: July 21, 2011, 03:27:22 PM »
let us regard MnCl2:

1 mole of MnCl2 contains how many moles of Cl?

Offline Vidya

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Re: How many moles of Cl2...?
« Reply #2 on: July 21, 2011, 09:33:57 PM »
You can use this ratio
 2 mole MnCl2 --- 5 mole Cl2

0.5714285714 molesMnCl2 -----??? Cl2

Offline vmelkon

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Re: How many moles of Cl2...?
« Reply #3 on: July 24, 2011, 10:41:36 AM »
You can use this ratio
 2 mole MnCl2 --- 5 mole Cl2

0.5714285714 molesMnCl2 -----??? Cl2

Don't you mean
2 moles MnCl2 has 2 moles Cl2

Offline Vidya

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Re: How many moles of Cl2...?
« Reply #4 on: July 24, 2011, 11:29:16 AM »
No
I mean when 2 moles of MnCl2 are formed at the same time 5 moles of Cl2 are also formed .This ratio we have taken from balanced reaction .

Offline adam456

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Re: How many moles of Cl2...?
« Reply #5 on: August 01, 2011, 05:58:07 PM »
You can use this simple formula

# of Moles = Mass in grams divided by Molar Mass

=5o divided by (cl x 2)

=50 divided by 71

=0.704 moles



Offline Dan

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Re: How many moles of Cl2...?
« Reply #6 on: August 01, 2011, 06:50:04 PM »
I'm trying to figure out how many moles of Cl2 are in the product side of this equation:
2HMnO4+ 14HCl --> 2MnCl2+ 5Cl2+ 8H2O
I know that the HCl is the limiting reactant and 2MnCl2 is 0.5714285714 moles, but how do I figure out how many moles are in the Cl2 part of the 2MnCl2???

There is some confusion here I think. If the question asks for how much Cl2 is formed, this is normally read to mean how much diatomic chlorine is formed. The Cl2 in MnCl2 is two chlorides, not diatomic chlorine.

You can use this simple formula

# of Moles = Mass in grams divided by Molar Mass

=5o divided by (cl x 2)

=50 divided by 71

=0.704 moles

How did you come to the conclusion that 50 g of Cl2 are formed? I think your answer is wrong.

Uma's approach is the simplest. If you know how much MnCl2 is formed, you can calculate how much Cl2 is formed using simple stoichiometry.
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