# 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?

Nov 12, 2015

Who cares about the chlorine? Grab the GOLD!

#### Explanation:

You have the equation:

$2 A u C {l}_{3} \rightarrow 2 A u + 3 C {l}_{2}$ (the which would probably rely on electrochemical reduction). There are $\frac{64.0 \cdot g}{303.33 \cdot g \cdot m o {l}^{- 1}}$ $=$ $0.211$ $m o l$ $A u C {l}_{3}$. By the stoichiometry of the reaction, you know that $\frac{3}{2}$ moles of chlorine ($C {l}_{2}$) result from each mole of $A u C {l}_{3}$. So $0.316$ moles of $C {l}_{2}$ will result. How many grams does this represent if the mass of molecular chlorine is $70.906 \cdot g \cdot m o {l}^{- 1}$? Approx. $21$ $g$?

A more realistic question would ask you to calculate the volume of such $C {l}_{2}$ 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.