Determine the mass of salt and the amount of water obtained by the interaction

| April 20, 2021 | education

Determine the mass of salt and the amount of water obtained by the interaction of copper (II) hydroxide with hydrochloric acid, if it is known that a solution weighing 365 g contains 10% acid.

Data: mр is the mass of the acid solution (mр = 365 g); ωHCl – mass fraction of pure acid (ωHCl = 0.1).

Const: MHCl – molar mass of hydrochloric acid (MHCl ≈ 36.5 g / mol); MCuCl2 – molar mass of copper chloride (MCuCl2 ≈ 134.5 g / mol).

1) The mass of pure acid: mHCl = mr * ωHCl = 365 * 0.1 = 36.5 g.

2) Amount of hydrochloric acid substance: νHCl = mHCl / MHCl = 36.5 / 36.5 = 1 mol.

3) Ur-tion of the reaction: Cu (OH) 2 (copper hydroxide) + 2HCl (hydrochloric acid) = CuCl2 (copper chloride) + 2H2O (water).

4) Amount of salt substance (copper chloride): νCuCl2 / νHCl = 1/2 and νCuCl2 = νHCl / 2 = 1/2 = 0.5 mol.

5) Mass of copper chloride: mCuCl2 = νCuCl2 * MCuCl2 = 0.5 * 134.5 = 67.25 g.

6) Amount of water substance: νН2О / νHCl = 2/2 and νН2О = νHCl = 1 mol.

Answer: The mass of salt (copper chloride) is 67.25 g, the amount of water is 1 mol.



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