Determine the mass of salt that is formed by the interaction of 30 g of ethanic acid with a solution containing 10 g of NaOH.

| August 2, 2021 | education

Data: mСН3СООH is the mass of ethanic (acetic) acid taken (mСН3СООH = 30 g); mNaOH is the mass of sodium hydroxide in solution (mNaOH = 10 g).

Const: MCH3COOH – molar mass of ethanic (acetic) acid (MCH3COOH ≈ 60 g / mol); MNaOH is the molar mass of sodium hydroxide (MNaOH ≈ 40 g / mol); MСН3СООNa – molar mass of sodium acetate (for the anhydrous form MСН3СООNa ≈ 82 g / mol).

1) The considered reaction: CH3COOH (ethanoic acid) + NaOH (sodium hydroxide) = CH3COONa (sodium acetate, salt formed) + H2O (water).

2) Amounts of substance:

– ethanic acid: νCH3COOH = mCH3COOH / MCH3COOH = 30/60 = 0.5 mol;

– sodium hydroxide: νNaOH = mNaOH / MNaOH = 10/40 = 0.25 mol (in deficit).

3) The amount of the substance of the formed sodium acetate: νСН3СООNa = νNaOH = 0.25 mol.

4) Mass of sodium acetate: mСН3СООNa = νСН3СООNa * MСН3СООNa = 0.25 * 82 = 20.5 g.

Answer: Sodium acetate was formed in an amount of 20.5 g.



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