Question

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 11.0 g of aluminum

Answer 1

Answer: 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.

Explanation:

To calculate the moles :

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

$\text{Moles of aluminium}=\frac{11.0g}{27g/mol}=0.407moles$

$2Al+3Cl_2\rightarrow 2AlCl_3$  

According to stoichiometry :

$Al$ is the limiting reagent as it limits the formation of product and $Cl_2$ is the excess reagent.

As 2 moles of $Al$ give = 2 moles of $AlCl_3$

Thus moles of $Al$ give =$\frac{2}{2}\times 0.407=0.407moles$  of $AlCl_3$

Thus 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.