Respuesta :
Answer: The amount of barium sulfate produced in the given reaction is 0.667 grams.
Explanation:
To calculate the number of moles from molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of barium chloride = 0.113 M
Volume of barium chloride = 25.34 mL = 0.02534 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol[/tex]
For the given chemical reaction:
[tex]BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)[/tex]
By Stoichiometry of the reaction:
1 mole of barium chloride is producing 1 mole of barium sulfate.
So, 0.00286 moles of barium chloride will produce = [tex]\frac{1}{1}\times 0.00286mol=0.00286mol[/tex] of barium sulfate.
Now, to calculate the mass of barium sulfate, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of barium sulfate = 233.38 g/mol
Moles of barium sulfate = 0.00286 moles
Putting values in above equation, we get:
[tex]0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g[/tex]
Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams
