<u>Answer:</u> The mass of barium sulfate produced is 1.45 grams.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of
= 1.5052 g
Molar mass of
= 244.26 g/mol
Putting values in equation 1, we get:

The chemical equation for the reaction of barium chloride and sulfuric acid follows:

As, sulfuric acid is present in excess, it is considered as an excess reagent.
Thus,
is considered as a limiting reagent because it limits the formation of product
By Stoichiometry of the reaction:
1 mole of
produces 1 mole of barium sulfate
So, 0.0062 moles of
will produce =
moles of barium sulfate
Now, calculating the mass of barium sulfate by using equation 1:
Molar mass of barium sulfate = 233.4 g/mol
Moles of barium sulfate = 0.0062 moles
Putting values in equation 1, we get:

Hence, the mass of barium sulfate produced is 1.45 grams.