Question

Найдите объём аммиака в литрах н.у. который образуется при взаимодействии избытка хлорида аммония и 9,25 г технического гидроксида кальция содержащего 20% примесей Ответ запишите в виде числа с точностью до сотых отделяет целую часть от дробной с помощью точки​

Answer 1

Ответ:

Объяснение:

2NH4Cl + Ca(OH)2 = 2NH3 + H2O + CaCl2

m(Ca(OH)2) без примесей = 9.25 г * 0.8 = 7.4 г

n(Ca(OH)2) = 7.4 г/74 г/моль = 0.1 моль

Химическое количество аммиака по уравнению реакции будет в 2 раза больше.

V(NH3) = 0.2 моль * 22.4 дм3/моль = 4.48 дм3

Answer 2

Calculating the Volume of Ammonia Produced

To calculate the volume of ammonia produced, we can use the balanced chemical equation for the reaction between excess ammonium chloride and 9.25g of technical grade calcium hydroxide containing 20% impurities. The balanced chemical equation for the reaction is:

Ca(OH)2 + 2NH4Cl → 2NH3 + CaCl2 + 2H2O

First, we need to calculate the moles of the limiting reactant, which is the technical grade calcium hydroxide.

The molar mass of Ca(OH)2 is 74.093 g/mol. The molar mass of NH4Cl is 53.491 g/mol.

Using the given mass of technical grade calcium hydroxide (9.25g) and the fact that it contains 20% impurities, we can calculate the moles of Ca(OH)2.

Moles of Ca(OH)2 = (mass of Ca(OH)2 / molar mass of Ca(OH)2) * (1 - 20%)

Next, we can use the moles of Ca(OH)2 to find the moles of NH3 produced, and then convert the moles of NH3 to volume at standard temperature and pressure (STP).

I will now calculate the moles of Ca(OH)2 and the volume of ammonia produced.