What is the mass of the precipitate obtained by the interaction of 10 g of BaCl2 with Na2SO4, if the yield = 80%?

The interaction of the sulfate ion with the barium ion is a qualitative reaction to the barium ion. As a result, we obtain an insoluble barium sulfate of a fine fraction, reminiscent of milk in solution.
Let’s write the reaction equation:
BaCl2 + Na2SO4 = BaSO4 + 2NaCl
To find the mass of the sediment, let’s make the proportion:
10 – x
208 – 233
Where: x = 10 * 233/208 = 11.2 grams
According to the condition of the problem, the reaction yield is 80%, that is, we will receive only 80% of the reaction product from the theoretically possible. We find the mass of the sediment which we get:
m = 11.2 * 0.8 = 8.96 grams.