NaOH + HNO3 = ? уравнение реакции К 39 мл раствора гидроксида натрия с массовой долей 20% и плотностью 1,2 г/мл добавили 58,66 г раствора азотной кислоты (NaOH + HNO3 = ?) с концентрацией 15%. Определите массу образовавшейся соли, запишите молекулярное и ионное уравнение реакции.

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Molecular equation:
NaOH + HNO3 = NaNO3 + H2O

Ionic equation:
Na+ + OH- + H+ + NO3- = Na+ + NO3- + H2O

First, let's calculate the moles of NaOH in the solution:
Volume of NaOH solution = 39 ml
Mass fraction of NaOH = 20%
Density of NaOH solution = 1.2 g/ml

Mass of NaOH = Volume x Density x Mass fraction
Mass of NaOH = 39 ml x 1.2 g/ml x 0.20
Mass of NaOH = 9.36 g

Moles of NaOH = Mass / Molar Mass
Moles of NaOH = 9.36 g / 40 g/mol
Moles of NaOH = 0.234 mol

Now, let's calculate the moles of HNO3 in the solution:
Mass of HNO3 = 58.66 g
Concentration of HNO3 = 15%

Moles of HNO3 = Mass / Molar Mass
Moles of HNO3 = 58.66 g / 63 g/mol (molar mass of HNO3)
Moles of HNO3 = 0.93 mol

Since NaOH and HNO3 react in a 1:1 ratio according to the balanced equation, the limiting reagent is NaOH. Therefore, all NaOH will react with HNO3.

The mass of the resulting salt, NaNO3, can be calculated using the moles of NaOH:
Mass of NaNO3 = Moles of NaOH x Molar Mass of NaNO3
Mass of NaNO3 = 0.234 mol x (23 + 14 + 48) g/mol
Mass of NaNO3 = 0.234 mol x 85 g/mol
Mass of NaNO3 = 19.89 g

Therefore, the mass of the resulting salt, NaNO3, is 19.89 g.