Question

Eading and

Answer 1

Answer: 25.8 g of $Cl_2$ will be produced from the decomposition of 73.4 g of $AuCl_3$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles$

The balanced chemical reaction is:

$2AuCl_3\rightarrow 2Au+3Cl_2$  

According to stoichiometry :

2 moles of $AuCl_3$ produce =  3 moles of $Cl_2$

Thus 0.242 moles of  will produce= $\frac{3}{2}\times 0.242=0.363mol$ of $Cl_2$

Mass of $Cl_2$= $moles\times {\text {Molar mass}}=0.363mol\times 71g/mol=25.8g$

Thus 25.8 g of $Cl_2$ will be produced from the decomposition of 73.4 g of $AuCl_3$