Answer: 25.8 g of [tex]Cl_2[/tex] will be produced from the decomposition of 73.4 g of [tex]AuCl_3[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles[/tex]
The balanced chemical reaction is:
[tex]2AuCl_3\rightarrow 2Au+3Cl_2[/tex]
According to stoichiometry :
2 moles of [tex]AuCl_3[/tex] produce = 3 moles of [tex]Cl_2[/tex]
Thus 0.242 moles of will produce= [tex]\frac{3}{2}\times 0.242=0.363mol[/tex] of [tex]Cl_2[/tex]
Mass of [tex]Cl_2[/tex]= [tex]moles\times {\text {Molar mass}}=0.363mol\times 71g/mol=25.8g[/tex]
Thus 25.8 g of [tex]Cl_2[/tex] will be produced from the decomposition of 73.4 g of [tex]AuCl_3[/tex]
