Question

Сколько миллилитров концентрированной соляной кислоты с плотностью 1,2 г/мл c массовой долей соляной кислоты 39% теоретически необходимо для взаимодействия с 0,1 моль перманганата калия? Какой объем хлора выделится при этом при температуре 0 °C и давлении 101,3 кПа?

Answer 1

16HCl + 2KMnO4 = 2MnCl2 + 5Cl2 + 8H2O + 2KCl
      16           2
        x           0.1
х = 16*0,1/2 = 0,8  моль.
масса кислоты = 0,8*36,5 = 29,2 г
масса 39%-ного раствора к-ты = 29,2/0,39 = 74,87 г
Объём 39-%-ного раствора = m/p = 74.87/1.2 = 62.39 мл

Answer 2

Calculation of the Volume of Concentrated Hydrochloric Acid Required

To calculate the volume of concentrated hydrochloric acid required to react with 0.1 mole of potassium permanganate, we need to use the balanced chemical equation for the reaction between hydrochloric acid (HCl) and potassium permanganate (KMnO4). The balanced equation is as follows:

8HCl + 2KMnO4 -> 2KCl + 2MnCl2 + 5Cl2 + 4H2O

From the balanced equation, we can see that 8 moles of hydrochloric acid react with 2 moles of potassium permanganate to produce 5 moles of chlorine gas. Therefore, the molar ratio between hydrochloric acid and potassium permanganate is 8:2.

Given that we have 0.1 mole of potassium permanganate, we can calculate the moles of hydrochloric acid required using the molar ratio:

0.1 mole KMnO4 * (8 moles HCl / 2 moles KMnO4) = 0.4 moles HCl

Now, we need to convert the moles of hydrochloric acid to volume using the concentration and density of the concentrated hydrochloric acid.

The concentration of the hydrochloric acid is given as 39% (mass/volume). This means that 100 mL of the concentrated hydrochloric acid contains 39 grams of hydrochloric acid.

The density of the concentrated hydrochloric acid is given as 1.2 g/mL.

To calculate the volume of the concentrated hydrochloric acid required, we can use the following formula:

Volume (mL) = (0.4 moles HCl * 36.5 g/mole) / (39 g/100 mL * 1.2 g/mL)

Let's calculate the volume of the concentrated hydrochloric acid required using the given values:

Volume (mL) = (0.4 * 36.5) / (39 * 1.2) = 0.391 mL

Therefore, approximately 0.391 mL of concentrated hydrochloric acid with a density of 1.2 g/mL and a mass fraction of 39% is theoretically required to react with 0.1 mole of potassium permanganate.

Calculation of the Volume of Chlorine Gas Produced

To calculate the volume of chlorine gas produced, we need to use the ideal gas law, which states that the volume of a gas is directly proportional to the number of moles of the gas, at constant temperature and pressure.

Given that the temperature is 0 °C and the pressure is 101.3 kPa, we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure (in kPa) - V is the volume (in liters) - n is the number of moles - R is the ideal gas constant (0.0821 L·atm/(mol·K)) - T is the temperature (in Kelvin)

We know that 5 moles of chlorine gas are produced from the reaction between 8 moles of hydrochloric acid and 2 moles of potassium permanganate.

Using the given values, we can calculate the volume of chlorine gas produced:

V = (n * R * T) / P

Let's calculate the volume of chlorine gas produced using the given values:

V = (5 * 0.0821 * 273) / 101.3 = 1.39 L

Therefore, approximately 1.39 liters of chlorine gas will be produced at a temperature of 0 °C and a pressure of 101.3 kPa.

Please note that these calculations are based on theoretical values and may vary in practice due to various factors such as reaction efficiency and experimental conditions.