Answer: 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of aluminium}=\frac{11.0g}{27g/mol}=0.407moles[/tex]
[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]
According to stoichiometry :
[tex]Al[/tex] is the limiting reagent as it limits the formation of product and [tex]Cl_2[/tex] is the excess reagent.
As 2 moles of [tex]Al[/tex] give = 2 moles of [tex]AlCl_3[/tex]
Thus moles of [tex]Al[/tex] give =[tex]\frac{2}{2}\times 0.407=0.407moles[/tex] of [tex]AlCl_3[/tex]
Thus 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.
