An excess of nitric acid solution was added to a solution of sodium hydroxide with a volume of 650 ml

| August 1, 2021 | education

An excess of nitric acid solution was added to a solution of sodium hydroxide with a volume of 650 ml, density 1.44 g / ml, with a mass fraction of 40%. Find the mass of the salt formed.

The reaction proceeds: NaOH + HNO3 = NaNO3 + H2O.
1) Calculate the mass of the caustic soda solution. To do this, multiply the volume of the solution by its density: 650 * 1.44 = 936g.
2) Calculate the mass of the hydroxide solute. To do this, we multiply the mass of the solution by the mass fraction of the substance and divide by one hundred percent: 936 * 40/100 = 374.4 g.
3) Molar mass of sodium hydroxide: 23 + 16 + 1 = 40.
4) The amount of sodium hydroxide substance: 374.4 / 40 = 9.36 mol.
5) According to the reaction equation, there is one mole of sodium nitrate for one mole of hydroxide. This means that the amount of nitrate substance is 9.36 mol.
6) The mass of salt is: 85 * 9.36 = 795.6 g – the answer.



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