Question

Помогите срочноВ лаборатории хлор получаютВ лаборатории хлор получают взаимодействием концентрированной соляной кислоты и оксида марганца (IV):HCl+MnO2=Cl2+MnCl2+H2OРасставьте коэффициенты в уравнении этой реакции, укажите окислитель и восстановитель и рассчитайте объем выделившегося хлора при вступлении в реакцию 3 моль оксида марганца (IV).СРОЧНО!!! Заранее спасибо)))

Answer 1

              3 моль   хл
4H⁺Cl⁻+Mn⁺⁴O₂⁻²=Cl₂°+Mn⁺²Cl₂⁻¹+2H₂⁺O⁻²
            1моль    22,4л
х = 22,4*3/1=67,2л

2Сl⁻-2e=Сl₂°  1 в-ль, ок-ние
Mn⁺⁴+2e=Mn⁺² 1 ок-ль, вос-ние

Answer 2

Balancing the Chemical Equation:

The balanced chemical equation for the reaction between concentrated hydrochloric acid and manganese(IV) oxide is as follows:

HCl + MnO2 → Cl2 + MnCl2 + H2O

To balance this equation, we need to ensure that the same number of each type of atom is present on both sides of the equation. Let's balance the equation:

4HCl + MnO2 → Cl2 + 2MnCl2 + 2H2O

Identifying the Oxidizing and Reducing Agents:

In this reaction, manganese undergoes a change in oxidation state from +4 to +2, indicating that it has been reduced. Therefore, manganese(IV) oxide (MnO2) is the oxidizing agent. Hydrochloric acid (HCl) is the reducing agent.

Calculating the Volume of Chlorine Gas Produced:

The molar ratio between manganese(IV) oxide and chlorine gas can be used to calculate the volume of chlorine gas produced when 3 moles of manganese(IV) oxide react.

From the balanced equation, we can see that 1 mole of manganese(IV) oxide produces 1 mole of chlorine gas. Therefore, 3 moles of manganese(IV) oxide will produce 3 moles of chlorine gas.

Using the ideal gas law, we can calculate the volume of the produced chlorine gas at standard temperature and pressure (STP), which is 0°C and 1 atmosphere pressure.

At STP, 1 mole of any gas occupies 22.4 liters. Therefore, 3 moles of chlorine gas will occupy:

3 moles * 22.4 L/mole = 67.2 liters

So, when 3 moles of manganese(IV) oxide react, 67.2 liters of chlorine gas will be produced.

I hope this helps! If you have any further questions or need additional assistance, feel free to ask.