Question

ПОЖАЛУЙСТА ПОМОГИТЕ! ЭТО ОЧЕНЬ СРОЧНО!!!Нейтрализуйте выданный вам раствор гидроксида натрия раствором азотной кислоты (в присутствии индикатора). Составьте уравнение химической реакции в молекулярном и ионном виде. Рассчитайте, хватит ли раствора азотной кислоты массой 60г. с массовой долей HNO3 , равной 10% , на нейтрализацию гидроксида натрия в растворе массой 70г. с массовой долей NaOH , равной 5%. *Вычислите массу и массовую долю соли, образовавшейся в результате реакции

Answer 1

Дано:

m(р-ра NaOH)=70 г

w(NaOH)=5%

m(р-ра HNO3)=60 г

w(HNO3)=10%

_______________

Хватит ли HNO3?

m(NaNO3)-?

w(NaNO3)-?

Решение:

1 Уравнение реакции в молекулярном виде:

NaOH + HNO3 -> NaNO3 + H2O

ПИ:

Na(+)+OH(-) + H(+)+NO3(-) -> Na(+)+NO3(-) + H2O

СИ:

OH(-) + H(+) -> H2O

2 m(NaOH)=70×(5/100)=3.5 г

3 m(HNO3)=60×(10/100)=6 г

4 n=m/M

M(NaOH)=23+16+1=40 г/моль

n(NaOH)=3.5/40=0.0875 моль

- недостаток

5 M(HNO3)=1+14+48=63 г/моль

n(HNO3)=6/63=0.0952 моль

- избыток

=> HNO3 хватит

6 По уравнению:

n(NaOH)=n(NaNO3)=0.0875 моль

7 m=n×M

M(NaNO3)=23+14+16×3=85 г/моль

m(NaNO3)=0.0875×85=7.438 г

8 w(NaNO3)=(7.438/130)×100=5.7%

Ответ: HNO3 хватит, 7.438 г, 5.7% соли.

Answer 2

Neutralization of Sodium Hydroxide with Nitric Acid

To neutralize the given solution of sodium hydroxide with nitric acid in the presence of an indicator, we can first write the balanced chemical equation for the reaction in both molecular and ionic forms. Then, we can calculate whether the given mass of nitric acid solution is sufficient to neutralize the given mass of sodium hydroxide solution.

Balanced Chemical Equation

The neutralization reaction between sodium hydroxide (NaOH) and nitric acid (HNO3) can be represented by the following balanced chemical equation:

Molecular Equation: 2NaOH + HNO3 → NaNO3 + H2O

Ionic Equation: 2Na⁺ + 2OH⁻ + 2H⁺ + NO3⁻ → 2Na⁺ + NO3⁻ + 2H2O

Calculation of Nitric Acid Mass

To calculate whether the given mass of nitric acid solution (60g, 10% HNO3) is sufficient to neutralize the given mass of sodium hydroxide solution (70g, 5% NaOH), we can use the concept of equivalent weights.

The equivalent weight of a substance is the mass that would supply or react with 1 mole of hydrogen ions (H⁺) or hydroxide ions (OH⁻). For nitric acid (HNO3), the equivalent weight is equal to its molar mass divided by the number of acidic hydrogen ions it can donate.

The molar mass of HNO3 is approximately 63.01 g/mol. Since it is a monoprotic acid (donates one hydrogen ion), the equivalent weight of HNO3 is also 63.01 g/mol.

For sodium hydroxide (NaOH), the equivalent weight is equal to its molar mass divided by the number of hydroxide ions it can donate. The molar mass of NaOH is approximately 40.00 g/mol. Since it is a monobasic base (donates one hydroxide ion), the equivalent weight of NaOH is 40.00 g/mol.

Now, we can calculate the number of moles of each substance present in the given masses of the solutions and determine the limiting reagent to find out if the given mass of nitric acid solution is sufficient to neutralize the given mass of sodium hydroxide solution.

Calculation of Salt Mass and Composition

After the neutralization reaction, the salt formed is sodium nitrate (NaNO3). We can calculate the mass and mass fraction of the salt formed using the stoichiometry of the reaction.

The molar mass of NaNO3 is approximately 85.00 g/mol. By using the stoichiometry of the reaction, we can calculate the mass of NaNO3 formed and then determine its mass fraction.

Let's proceed with the calculations to determine the sufficiency of the given nitric acid solution and the mass and mass fraction of the salt formed.