To completely neutralize 80 g of a 4.5% nitric acid solution, 40 g of a potassium hydroxide solution containing 10%

| May 21, 2021 | education

To completely neutralize 80 g of a 4.5% nitric acid solution, 40 g of a potassium hydroxide solution containing 10% impurities were required. Find the volume of water.

HNO3 + KOH = KNO3 + H2O.
1) The mass of the dissolved substance of nitric acid: 80 * 4.5 / 100 = 3.6 g.
2) Molar mass of nitric acid: 1 + 14 + 16 * 3 = 63.
3) Amount of nitric acid substance: 3.6 / 63 = 0.06 mol.
4) The mass of impurities in potassium hydroxide: 40 * 10/100 = 4g.
5) The mass of the pure substance of potassium hydroxide: 40 – 4 = 36g.
6) Molar mass of potassium hydroxide: 39 + 1 + 16 = 56.
7) Amount of potassium hydroxide substance: 36/56 = 0.64 mol.
Lack of nitric acid. The calculation is carried out on it.
8) The amount of water substance is equal to: 0.06 mol.
9) The volume of water is: 0.06 * 22.4 = 1.344l – the answer.



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