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)

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:

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