# Write the ionic equations of the reactions between dissolved in water A) copper sulfate (2)

Write the ionic equations of the reactions between dissolved in water A) copper sulfate (2) and barium chloride b) aluminum sulfate and barium chloride c) sodium sulfate and barium nitrate.

1. Copper (II) sulfate – CuSO4, barium chloride – BaCl2
CuSO4 + BaCl2 = CuCl2 + BaSO4 (sediment)
In solution: Cu2 + + SO4 2- + Ba2 + + 2Cl- = Cu2 + + 2Cl- + BaSO4 (sediment)
Let’s remove the ions of the same name in the left and right sides of the equation
Abbreviated form: Ba2 + + SO4 2- = BaSO4 (sediment)
2.Aluminum sulfate Al2 (SO4) 3, barium chloride – BaCl2
Al2 (SO4) 3 + 3BaCl2 = 2AlCl3 + 3BaSO4 (sediment)
2Al3 + + 3SO4 2- + 3Ba2 + + 6Cl- = 2Al3 + + 6Cl- + 3BaSO4 (precipitate)
Let’s remove the ions of the same name in the left and right sides of the equation, divide the coefficients of the remaining ions by 3:
Ba2 + + SO4 2- = BaSO4 (sediment) – an abbreviated form of the ionic equation
3. Sodium sulfate – Na2SO4, barium nitrate – Ba (NO3) 2
Na2SO4 + Ba (NO3) 2 = 2NaNO3 + BaSO4 (sediment)
2Na + + SO4 2- + Ba2 + + 2NO3- = 2Na + + 2NO3- + BaSO4 (sediment)
Let’s remove the ions of the same name in the left and right sides of the equation
Abbreviated form: Ba2 + + SO4 2- = BaSO4 (sediment)

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