How many grams of chlorine can be liberated from the decomposition of 64.0g of AuCl3 by the following reaction: 2 AuCl3 --> 2 Au + 3Cl2?

1 Answer
Nov 12, 2015

Answer:

Who cares about the chlorine? Grab the GOLD!

Explanation:

You have the equation:

#2AuCl_3 rarr 2Au + 3Cl_2# (the which would probably rely on electrochemical reduction). There are #(64.0*g)/(303.33*g*mol^(-1))# #=# #0.211# #mol# #AuCl_3#. By the stoichiometry of the reaction, you know that #3/2# moles of chlorine (#Cl_2#) result from each mole of #AuCl_3#. So #0.316# moles of #Cl_2# will result. How many grams does this represent if the mass of molecular chlorine is #70.906*g*mol^(-1)#? Approx. #21# #g#?

A more realistic question would ask you to calculate the volume of such #Cl_2# at standard temperature and pressure.

This seems to be a very expensive way to make chlorine. Would you throw the mixture down the sink when you're finished? I've seen it done.