# Mixed solutions containing 31.2 g of barium chloride and 136.8 g of aluminum sulfate.

**Mixed solutions containing 31.2 g of barium chloride and 136.8 g of aluminum sulfate. Determine which of the substances is in excess. Calculate the mass of this substance remaining in the solution after the reaction.**

Given:

m (BaCl2) = 31.2 g

m (Al2 (SO4) 3) = 136.8 g

To find:

in excess -?

m (the remainder of the islands in excess) -?

Solution:

1) 3BaCl2 + Al2 (SO4) 3 => 2AlCl3 + 3BaSO4 ↓;

2) n (BaCl2) = m (BaCl2) / M (BaCl2) = 31.2 / 208 = 0.15 mol;

3) n (Al2 (SO4) 3) = m (Al2 (SO4) 3) / M (Al2 (SO4) 3) = 136.8 / 342 = 0.4 mol;

4) Al2 (SO4) 3 – in excess;

5) n react. (Al2 (SO4) 3) = n (BaCl2) / 3 = 0.15 / 3 = 0.05 mol;

6) n rest. (Al2 (SO4) 3) = n (Al2 (SO4) 3) – n reag. (Al2 (SO4) 3) = 0.4 – 0.05 = 0.35 mol;

7) m rest. (Al2 (SO4) 3) = n rest. (Al2 (SO4) 3) * M (Al2 (SO4) 3) = 0.35 * 342 = 119.7.

Answer: The mass of the remaining Al2 (SO4) 3 is 119.7.