[horsegurl221]

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???

[DevaDevil]

let us regard MnCl₂:

1 mole of MnCl₂ contains how many moles of Cl?

[Vidya]

You can use this ratio
 2 mole MnCl2 --- 5 mole Cl2

0.5714285714 molesMnCl2 -----??? Cl2

[vmelkon]

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

[Vidya]

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 .

[adam456]

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

[Dan]

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.