In water, 1 mol of aluminum chloride, 3 mol of sodium hydroxide, 2 mol of barium chloride, 1 mol

In water, 1 mol of aluminum chloride, 3 mol of sodium hydroxide, 2 mol of barium chloride, 1 mol of potassium sulfate, 2 mol of silver nitrate were simultaneously dissolved. What ions are present in the solution after the separation of the precipitate. Calculate the amount of substance of each ion

1. AlCl3 + 3NaOH → Al (OH) 3 ↓ + 3NaCl. All aluminum chloride and all alkali are consumed, after the reaction, 3 mol of Na + ion and 3 Cl- ion will remain in the system;

2. BaCl2 + K2SO4 → BaSO4 ↓ + 2KCl. Of 2 mol of barium chloride, only half reacts – 1 mol, as a result, this reaction will give 2 mol of K + ion and 2 mol of Cl-ion, and from the remainder of 1 mol of BaCl2 – 1 mol of Ba2 + ion and 2 mol of Cl- ion. Part of Cl- ions will precipitate after reaction with AgNO3;

3. AgNO3 + Cl- → AgCl ↓ + NO3-. Thus, silver will bind 2 mol of the Cl- ion and supply 2 mol of the NO3- ion to the system. Let’s write down all the ions remaining in the solution after the reaction:

4.3Na + + 3Cl- + 2K + + 2Cl- + Ba2 + + 2Cl- – 2Cl- + 2NO3- = 3Na + + 5Cl- + 2K + + Ba2 + + 2NO3-.

Answer: after separation of the precipitate, the solution contains 3 mol Na +, 5 mol Cl-, 2 mol K +, 1 mol Ba2 + and 2 mol NO3-.