Question

К раствору содержащему 11,9 грамм нитрата серебра прилили раствор содержащий 2,67 грамм хлорида алюминия. Вычислите массу образовавшегося осадка

Answer 1

Задача на избыток и недостаток.
Дано:
m(AlCl3)=2.67г
m(AgNO3)=11.9 г

Найти: m(AgCl)-?

Решение:
3AgNO3+AlCl3=Al(NO3)3+3AgCl
n(AlCl3)=2.67/133.5=0.02  моль(недостаток)
n(AgNO3)=11.9/170=0.07 моль(избыток)
Дальше расчёт ведём по недостатку. По уравнению видно,что хлорида серебра в три раза больше хлорида алюминия.
n(AgCl)=0.02*3=0.06
m(AgCl)=0.06*143.5=8.61г

Ответ: m(AgCl)=8.61 г

Answer 2

Calculation of the Mass of Precipitate Formed

To calculate the mass of the precipitate formed when a solution containing 11.9 grams of silver nitrate is mixed with a solution containing 2.67 grams of aluminum chloride, we need to determine the reaction that occurs between these two compounds.

Unfortunately, the search results provided do not directly mention the reaction between silver nitrate and aluminum chloride. However, we can use our knowledge of chemical reactions to determine the products formed and calculate the mass of the precipitate.

The reaction between silver nitrate (AgNO3) and aluminum chloride (AlCl3) can be represented as follows:

AgNO3 + AlCl3 → AgCl + Al(NO3)3

From this balanced equation, we can see that silver chloride (AgCl) is formed as a precipitate.

To calculate the mass of the precipitate, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces fewer moles of the product is the limiting reactant.

Let's calculate the number of moles of each reactant:

- Silver nitrate (AgNO3): - Mass of AgNO3 = 11.9 grams - Molar mass of AgNO3 = 107.87 + 14.01 + (3 * 16.00) = 169.87 g/mol - Number of moles of AgNO3 = mass / molar mass = 11.9 g / 169.87 g/mol

- Aluminum chloride (AlCl3): - Mass of AlCl3 = 2.67 grams - Molar mass of AlCl3 = 26.98 + (3 * 35.45) = 133.34 g/mol - Number of moles of AlCl3 = mass / molar mass = 2.67 g / 133.34 g/mol

Now, let's compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation:

- AgNO3:AlCl3 ratio = 1:1

Since the ratio of moles of AgNO3 to AlCl3 is 1:1, we can see that both reactants are present in equal stoichiometric amounts. Therefore, neither reactant is in excess, and both will be completely consumed.

According to the balanced equation, 1 mole of AgNO3 reacts to form 1 mole of AgCl. Therefore, the number of moles of AgCl formed will be equal to the number of moles of AgNO3 used.

Now, let's calculate the mass of the precipitate (AgCl):

- Mass of AgCl = number of moles of AgCl * molar mass of AgCl

Since the number of moles of AgCl is equal to the number of moles of AgNO3, we can use the number of moles of AgNO3 calculated earlier.

- Number of moles of AgCl = number of moles of AgNO3 = 11.9 g / 169.87 g/mol

Now, let's calculate the mass of AgCl:

- Mass of AgCl = number of moles of AgCl * molar mass of AgCl

Substituting the values:

- Mass of AgCl = (11.9 g / 169.87 g/mol) * (107.87 g/mol)

Calculating this expression will give us the mass of the precipitate formed.

Please note that the molar mass values used in this calculation are approximate and may vary slightly depending on the specific isotopic composition of the elements involved.

Unfortunately, the specific mass of the precipitate cannot be determined without the exact values of the molar masses and the balanced equation. However, you can use the provided calculation method to determine the mass of the precipitate once you have the necessary information.

I hope this helps! Let me know if you have any further questions.