Question

How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2SO4(aq) → BaSO4 (s) + 2NaCl (aq)

Answer 1

Answer: The amount of barium sulfate produced in the given reaction is 0.667 grams.

Step-by-step explanation:

To calculate the number of moles from molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

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:

`0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol`

For the given chemical reaction:

`BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)`

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 =
`(1)/(1)* 0.00286mol=0.00286mol` of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

`0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g`

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams