Question

Смешали два раствора. Первый раствор содержал 1 моль хлорида аммония, а другой - 2 моль нитрата серебра. Найдите массу (в граммах) выпавшего осадка и запишите.

Answer 1

Ответ:

Надеюсь, что можно разобраться

Объяснение:

Answer 2

Calculation of the Mass of Precipitate

To find the mass of the precipitate formed when two solutions are mixed, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.

In this case, we have two solutions: one containing 1 mole of ammonium chloride (NH4Cl) and the other containing 2 moles of silver nitrate (AgNO3).

The balanced chemical equation for the reaction between ammonium chloride and silver nitrate is:

NH4Cl + AgNO3 -> AgCl + NH4NO3

From the equation, we can see that 1 mole of ammonium chloride reacts with 1 mole of silver nitrate to form 1 mole of silver chloride (AgCl) and 1 mole of ammonium nitrate (NH4NO3).

Since we have 1 mole of ammonium chloride and 2 moles of silver nitrate, the silver nitrate is in excess. Therefore, ammonium chloride is the limiting reactant.

To calculate the mass of the precipitate (silver chloride), we need to know the molar mass of silver chloride. The molar mass of AgCl is approximately 143.32 g/mol.

Now, let's calculate the mass of the precipitate:

1. Calculate the number of moles of ammonium chloride: - Number of moles = 1 mole

2. Calculate the number of moles of silver chloride formed: - From the balanced equation, we know that 1 mole of ammonium chloride reacts to form 1 mole of silver chloride. - Therefore, the number of moles of silver chloride formed is also 1 mole.

3. Calculate the mass of the precipitate (silver chloride): - Mass = Number of moles * Molar mass - Mass = 1 mole * 143.32 g/mol - Mass = 143.32 g

Therefore, the mass of the precipitate (silver chloride) formed is 143.32 grams.