In an experiment,0.197 g of gold metal reacts to yield 0.303 g of gold(III)chloride.Using the conservation of mass law,predict the mass of reacting chlorine gas. 2 Au(s)+ 3 Cl₂(g) 2 AuCl3(s)

Question

In an experiment,0.197 g of gold metal reacts to yield 0.303 g of gold(III)chloride.Using the conservation of mass law,predict the mass of reacting chlorine gas. 2 Au(s)+ 3 Cl₂(g)  2 AuCl3(s)

Quizplus · Accepted Answer

First, we need to use stoichiometry to determine the theoretical yield of AuCl3 that can be produced from 0.197 g of Au. 
Molar mass of Au = 196.97 g/mol 
0.197 g Au X (1 mol Au/196.97 g Au) X (2 mol AuCl3/2 mol Au) X (303.33 g AuCl3/mol AuCl3) = 0.303 g AuCl3 
So, 0.303 g of AuCl3 is produced. 
To determine the mass of Cl2 that reacted, we can use the mole ratio between AuCl3 and Cl2. 
From the balanced chemical equation, we can see that: 
2 moles of AuCl3 react with 3 moles of Cl2 
So, 0.303 g of AuCl3 reacts with (3/2) X (0.303 g) = 0.455 g of Cl2. 
However, we need to check if Cl2 is the limiting reagent. 
Molar mass of Cl2 = 70.90 g/mol 
0.455 g Cl2 X (1 mol Cl2/70.90 g Cl2) X (2 mol AuCl3/3 mol Cl2) X (303.33 g AuCl3/mol AuCl3) = 0.106 g AuCl3 
This is less than the actual yield of 0.303 g AuCl3, so Cl2 is not limiting. 
Therefore, the mass of Cl2 that reacted is 0.455 g.

In an Experiment,0