Question

1) Нагретый порошок железа сгорает в хлоре. В результате образуется хлорид железа (III) FeCl3, который используется для травления плат в электротехнике. Сгоранию железа в хлоре соответствует схема реакции: Fe+Cl2--->FeCl3Преобразуйте данную схему в уравнение реакции. Рассчитайте массу железа, которая потребуется для получения 381 грамма хлорида железа (III).2) Вычислите массу оксида алюминия, который образовался в результате сгорания 2,7 грамма порошка алюминия.

Answer 1

1) Запишем уравнение реакции: 2Fe + 3Cl2 = 2FeCl3
Находим количество FeCl3:
v(FeCl3) = 381 г : 162,5 г/моль = 2,345 моль.
Количество железа, по уравнению реакции, такое же, т.е. v(Fe) = v(FeCl3) = 2,345 моль.
Находим массу железа:
m(Fe) = 2,345 моль * 56 г/моль = 131,32 г.

2) Запишем уравнение реакции: 4Al + 3O2 = 2Al2O3
Находим количество алюминия:
v(Al) = 2,7 г : 27 г/моль = 0,1 моль.
Количество оксида алюминия, по уравнению реакции, в два раза меньше, т.е. v(Al2O3) = 0,05 моль.
Находим массу оксида алюминия:
m(Al2O3) = 0,05 моль * 102 г/моль = 5,1 г.

Answer 2

Reaction Equation for the Combustion of Iron in Chlorine

The reaction equation for the combustion of iron in chlorine can be represented as follows:

2Fe + 3Cl2 → 2FeCl3

Calculation of Iron Mass Required to Obtain 381 grams of Iron(III) Chloride

To calculate the mass of iron required to obtain 381 grams of iron(III) chloride (FeCl3), we can use the molar mass of FeCl3 and the stoichiometry of the reaction.

The molar mass of FeCl3 can be calculated as follows: - Atomic mass of iron (Fe) = 55.845 g/mol - Atomic mass of chlorine (Cl) = 35.453 g/mol - Molar mass of FeCl3 = (55.845 g/mol) + 3*(35.453 g/mol) = 162.205 g/mol

Using the stoichiometry of the reaction, we can set up the following proportion: - 2 moles of Fe react to form 2 moles of FeCl3 - X moles of Fe react to form 381 grams of FeCl3

By solving this proportion, we can find the moles of Fe required: - (2 moles Fe / 2 moles FeCl3) = (X moles Fe / 381 grams FeCl3)

Simplifying the proportion: - X moles Fe = (2 moles Fe / 2 moles FeCl3) * 381 grams FeCl3

Finally, we can calculate the mass of iron required: - Mass of iron = X moles Fe * molar mass of Fe

Let's calculate the mass of iron required:

Step 1: Calculate the moles of FeCl3: - Moles of FeCl3 = (381 grams FeCl3) / (162.205 g/mol) = 2.349 moles FeCl3

Step 2: Calculate the moles of Fe: - Moles of Fe = (2 moles Fe / 2 moles FeCl3) * 2.349 moles FeCl3 = 2.349 moles Fe

Step 3: Calculate the mass of iron: - Mass of iron = (2.349 moles Fe) * (55.845 g/mol) = 131.2 grams of iron

Therefore, approximately 131.2 grams of iron would be required to obtain 381 grams of iron(III) chloride.

Calculation of Aluminum Oxide Mass Formed from the Combustion of Aluminum

To calculate the mass of aluminum oxide formed from the combustion of 2.7 grams of aluminum, we need to consider the balanced chemical equation for the reaction.

The balanced chemical equation for the combustion of aluminum can be represented as follows:

4Al + 3O2 → 2Al2O3

According to the stoichiometry of the reaction, 4 moles of aluminum react to form 2 moles of aluminum oxide.

Step 1: Calculate the moles of aluminum: - Moles of aluminum = (2.7 grams Al) / (26.982 g/mol) = 0.1 moles Al

Step 2: Calculate the moles of aluminum oxide: - Moles of aluminum oxide = (2 moles Al2O3 / 4 moles Al) * 0.1 moles Al = 0.05 moles Al2O3

Step 3: Calculate the mass of aluminum oxide: - Mass of aluminum oxide = (0.05 moles Al2O3) * (101.96 g/mol) = 5.098 grams of aluminum oxide

Therefore, approximately 5.098 grams of aluminum oxide would be formed from the combustion of 2.7 grams of aluminum.