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<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.
<u>Explanation:</u>
To calculate the number of moles from molarity, we use the equation:
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:
For the given chemical reaction:
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 = of barium sulfate.
Now, to calculate the mass of barium sulfate, we use the equation:
Molar mass of barium sulfate = 233.38 g/mol
Moles of barium sulfate = 0.00286 moles
Putting values in above equation, we get:
Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams