Calculate the mass of the precipitate formed by adding an excess of potassium

| August 11, 2021 | education

Calculate the mass of the precipitate formed by adding an excess of potassium hydroxide to 19 g of magnesium chloride solution with a mass fraction of 5% salt.

Data: mr-ra is the mass of the magnesium chloride solution (mr-ra = 19 g); ωMgCl2 – mass fraction of magnesium chloride (ωMgCl2 = 5% = 0.05).

Const: MMgCl2 – molar mass of magnesium chloride (MMgCl2 = 95.211 g / mol ≈ 95 g / mol); МMg (ОН) 2 – molar mass of magnesium hydroxide (МMg (ОН) 2 = 58.35 g / mol ≈ 58 g / mol).

1) The mass of magnesium chloride in the taken solution: mMgCl2 = ωMgCl2 * mr-ra = 0.05 * 19 = 0.95 g.

2) The ur-tion of the reaction: Mg (OH) 2 (magnesium hydroxide) ↓ + 2KSl (potassium chloride) = 2KON (potassium hydroxide) + MgCl2 (magnesium chloride).

3) Amount of magnesium chloride substance: νMgCl2 = mMgCl2 / МMgCl2 = 0.95 / 95 = 0.01 mol.

4) Amount of magnesium hydroxide substance: νMg (OH) 2 / νMgCl2 = 1/1 and νMg (OH) 2 = νMgCl2 = 0.01 mol.

5) The mass of the precipitate formed: mMg (OH) 2 = νMg (OH) 2 + MMg (OH) 2 = 0.01 * 58 = 0.58 g.

Answer: The mass of the resulting precipitate (magnesium hydroxide) is 0.58 g.



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