<u>Answer:</u> The theoretical yield of barium sulfate is 50.9 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For sodium sulfate:</u>
Given mass of sodium sulfate = 32.4 g
Molar mass of sodium sulfate = 142 g/mol
Putting values in equation 1, we get:
- <u>For barium chloride:</u>
Given mass of barium chloride = 65.3 g
Molar mass of barium chloride = 208.23 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of barium chloride and sodium sulfate follows:
By Stoichiometry of the reaction:
1 mole of sodium sulfate reacts with 1 mole of barium chloride
So, 0.228 moles of sodium sulfate will react with = of barium chloride
As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.
Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sodium sulfate produces 1 mole of barium sulfate.
So, 0.228 moles of sodium sulfate will produce = of barium sulfate
Now, calculating the mass of barium sulfate from equation 1, we get:
Molar mass of barium sulfate = 233.4 g/mol
Moles of barium sulfate = 0.228 moles
Putting values in equation 1, we get:
Hence, the theoretical yield of barium sulfate is 50.9 grams