Question

Be sure to answer all parts. The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2AlCl3(s) Assume that 0.80 g Al is mixed with 0.23 g Cl2. (a) What is the limiting reactant? Cl2 Al (b) What is the maximum amount of AlCl3, in grams, that can be produced? g AlCl3

Answer 1

Answer: a) $Cl_2$ is the limiting reagent

b)  0.27 g of $AlCl_3$ will be produced.

Explanation:

To calculate the moles :

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

$\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles$

$\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles$

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

According to stoichiometry :

3 moles of $Cl_2$ require = 2 moles of $Al$

Thus 0.003 moles of $Cl_2$ will require=$\frac{2}{3}\times 0.003=0.002moles$  of $Al$

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

b) As 3 moles of $Cl_2$ give = 2 moles of $AlCl_3$

Thus 0.003 moles of $Cl_2$ give =$\frac{2}{3}\times 0.003=0.002moles$  of $AlCl_3$

Mass of $AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g$

Thus 0.27 g of $AlCl_3$ will be produced.