Question

Aluminum sulfate reacts with barium chloride to form the insoluble compound, barium sulfate. The reaction proceeds according to the balanced equation below:
1Al2(SO4)3 + 3BaCl2 → 2AlCl3 + 3BaSO4

Marie reacts 150 g of aluminum sulfate with 200 g of barium chloride in order to produce insoluble barium sulfate for her crystallography studies. Determine the limiting and excess reactants for this reaction.

Molar mass aluminum sulfate: 342.15 g/mol
Molar mass barium chloride: 208.23 g/mol
Molar mass barium sulfate: 233.38 g/mol

Answer 1

Answer: $BaCl_2$ is the limiting reagent and $Al_2(SO_4)_3$ is the excess reagent.

Explanation:

To calculate the moles :

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

$\text{Moles of aluminium sulphate}=\frac{150g}{342.15g/mol}=0.438moles$

$\text{Moles of barium chloride}=\frac{200g}{208.23g/mol}=0.960moles$

The balanced chemical reaction is:

$Al_2(SO_4)_3+3BaCl_2\rightarrow 2AlCl_3+3BaSO_4$  

According to stoichiometry :

3 moles of $BaCl_2$ require = 1 mole of $Al_2(SO_4)_3$

Thus 0.960 moles of $BaCl_2$ will require=$\frac{1}{3}\times 0.960=0.320moles$  of $Al_2(SO_4)_3$

Thus $BaCl_2$ is the limiting reagent as it limits the formation of product and $Al_2(SO_4)_3$ is the excess reagent as it is left.