Question

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?

Answer 1

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.