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 Cl

_{2} is formed, this is normally read to mean how much diatomic chlorine is formed. The Cl

_{2} in MnCl

_{2} 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 Cl

_{2} are formed? I think your answer is wrong.

Uma's approach is the simplest. If you know how much MnCl

_{2} is formed, you can calculate how much Cl

_{2} is formed using simple stoichiometry.