Answer: 1. [tex]2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu[/tex]
2. 3 moles of [tex]CuCl_2[/tex] : 2 moles of [tex]Al[/tex]
3. 0.33 moles of [tex]CuCl_2[/tex] : 0.92 moles of [tex]Al[/tex]
4. [tex]CuCl_2[/tex] is the limiting reagent and [tex]Al[/tex] is the excess reagent.
5. Theoretical yield of [tex]AlCl_3[/tex] is 29.3 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles[/tex]
[tex]\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles[/tex]
The balanced chemical equation is:
[tex]2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu[/tex]
According to stoichiometry :
3 moles of [tex]CuCl_2[/tex] require = 2 moles of [tex]Al[/tex]
Thus 0.33 moles of [tex]CuCl_2[/tex] will require=[tex]\frac{2}{3}\times 0.33=0.22moles[/tex] of [tex]Al[/tex]
Thus [tex]CuCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.
As 3 moles of [tex]CuCl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]
Thus 0.33 moles of [tex]CuCl_2[/tex] give =[tex]\frac{2}{3}\times 0.33=0.22moles[/tex] of [tex]AlCl_3[/tex]
Theoretical yield of [tex]AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3[/tex]
Thus 29.3 g of aluminium chloride is formed.