3. If 90.0 grams of barium chloride react with sulfuric acid, how many
grams of barium sulfate are produced?
BaCl2 + H250.-- BaSO4 + 2HCI

The answer to your question is 100.8 g

Explanation:

Data

  mass of Barium chloride = 90 g

  mass of Barium sulfate = ?

Balanced chemical reaction

    BaCl₂  +  H₂SO₄   ⇒    BaSO₄  +  2HCl

Process

1.- Calculate the molar mass of BaCl₂ and BaSO₄

BaCl₂ = 137.3 + (2 x 35.5) = 208.3g

BaSO₄ = 137.3 + 32 + (16 x 4) = 233.3g

2.- Use proportions to find the mass of BaSO₄

              208.3 g of BaCl₂ 233.3 g of BaSO₄

                90 g of BaCl₂       x

                              x = (90 x 233.3) / 208.3

                              x = 20997/ 208.3

                              x = 100.8 g

3.- Conclusion

There will be produced 100.8 g of BaSO₄          

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).

the answer is c. nacl