Respuesta :
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 :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of barium chloride}=\frac{76.0}{208g/mol}=0.365moles[/tex]
[tex]\text{Moles of potassium sulphate}=\frac{67.0}{174g/mol}=0.385moles[/tex]
[tex]BaCl_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KCl(aq)[/tex]
According to stoichiometry :
1 mole of [tex]BaCl_2[/tex] require 1 mole of [tex]K_2SO_4[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] will require=[tex]\frac{1}{1}\times 0.365=0.365moles[/tex] of [tex]K_2SO_4[/tex]
Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]K_2SO_4[/tex] is the excess reagent.
As 1 moles of [tex]BaCl_2[/tex] give = 1 moles of [tex]BaSO_4[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{1}{1}\times 0.365=0.365moles[/tex] of [tex]BaSO_4[/tex]
As 1 moles of [tex]BaCl_2[/tex] give = 2 moles of [tex]KCl[/tex]
Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{2}{1}\times 0.365=0.730moles[/tex] of [tex]KCl[/tex]
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.
