Question

For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate. barium chloride (aq) potassium sulfate (aq) barium sulfate (s) potassium chloride (aq)

Answer 1

For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.

The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.

Answer: a) The reaction consumes 0.365 moles of barium chloride.

b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

Explanation:

To calculate the moles :

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

$\text{Moles of barium chloride}=\frac{76.0}{208g/mol}=0.365moles$

$\text{Moles of potassium sulphate}=\frac{67.0}{174g/mol}=0.385moles$

$BaCl_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KCl(aq)$

According to stoichiometry :

1 mole of $BaCl_2$ require 1 mole of $K_2SO_4$

Thus 0.365 moles of $BaCl_2$ will require=$\frac{1}{1}\times 0.365=0.365moles$  of $K_2SO_4$

Thus $BaCl_2$ is the limiting reagent as it limits the formation of product and $K_2SO_4$ is the excess reagent.

As 1 moles of $BaCl_2$ give = 1 moles of $BaSO_4$

Thus 0.365 moles of $BaCl_2$ give =$\frac{1}{1}\times 0.365=0.365moles$  of $BaSO_4$

As 1 moles of $BaCl_2$ give = 2 moles of $KCl$

Thus 0.365 moles of $BaCl_2$ give =$\frac{2}{1}\times 0.365=0.730moles$  of $KCl$

Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.