Question

Вычислите, сколько объемов хлора можно получить при взаимодействии 43,5г оксида марганца с концентрированной соляной кислотой

Answer 1

Решение:

1).MnO2+4HCl = MnCl+2H20+Cl2

Кол-ва диоксида марганца и хлора будут одинаковыми то есть 0,5 моль.

V=Vm*n, V(Cl2) = 0,5 моль * 22,4 л/моль = 11,2 л

 Ответ: V(Cl2)= 11,2 л

Answer 2

Уравнение данной реакции: MnO2 + 4HCl = MnCl2 + 2H20 + Cl2

Количества веществ диоксида марганца и хлора равны n = 0,5 моль.

Тогда V(Cl2) = 0,5 моль * 22,4 л/моль = 11,2 л

Answer 3

Calculation of Chlorine Volume

To calculate the volume of chlorine that can be obtained from the reaction between 43.5g of manganese oxide and concentrated hydrochloric acid, we need to determine the stoichiometry of the reaction and use the molar masses of the reactants.

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

MnO + 4HCl → MnCl2 + 2H2O + Cl2

From the balanced equation, we can see that 1 mole of manganese oxide reacts with 4 moles of hydrochloric acid to produce 1 mole of chlorine gas.

Now, let's calculate the number of moles of manganese oxide:

Number of moles of MnO = Mass of MnO / Molar mass of MnO

The molar mass of manganese oxide (MnO) is 70.94 g/mol.

Number of moles of MnO = 43.5 g / 70.94 g/mol

To calculate the molar mass of MnO, we need to know the oxidation state of manganese in the oxide. Please provide the oxidation state of manganese in the oxide so that we can proceed with the calculation.