Answer: a) [tex]Cl_2[/tex] is the limiting reagent
b) 0.27 g of [tex]AlCl_3[/tex] will be produced.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles[/tex]
[tex]\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles[/tex]
[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]
According to stoichiometry :
3 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]Al[/tex]
Thus 0.003 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{3}\times 0.003=0.002moles[/tex] of [tex]Al[/tex]
Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.
b) As 3 moles of [tex]Cl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]
Thus 0.003 moles of [tex]Cl_2[/tex] give =[tex]\frac{2}{3}\times 0.003=0.002moles[/tex] of [tex]AlCl_3[/tex]
Mass of [tex]AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g[/tex]
Thus 0.27 g of [tex]AlCl_3[/tex] will be produced.