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

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kobenhavn kobenhavn · Answer 1 · 2021-03-27T03:54:22+01:00

Answer: [tex]BaCl_2[/tex] is the limiting reagent and [tex]Al_2(SO_4)_3[/tex] is the excess reagent.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]

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

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

The balanced chemical reaction is:

[tex]Al_2(SO_4)_3+3BaCl_2\rightarrow 2AlCl_3+3BaSO_4[/tex]

According to stoichiometry :

3 moles of [tex]BaCl_2[/tex] require = 1 mole of [tex]Al_2(SO_4)_3[/tex]

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

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