Question

Решите задачу: Определите массу оксида алюминия, которая может быть получена при полном сгорании 81 г алюминия в кислороде.

Answer 1

Уравнение реакции:

3 моль    х моль
4Al       + 3O₂         = 2Al₂O₃
4 моль    3 моль

Зная массу и молярную массу (=27, см. Периодическую таблицу), можем найти количество вещества алюминия:
n = m/M
n Al = 81 г / 27 г/моль = 3 моль

Решаем пропорцию, в ответе получаем количество кислорода:
х = 3*3/4
х = 2,25
Один моль любого газа при н.у. занимает объем 22,4 л. Следовательно,
22,4 л * 2,25 моль = 50,4 л
Ответ: 50,4 л.

Answer 2

Задача: Определение массы оксида алюминия при полном сгорании алюминия в кислороде

Для решения данной задачи, мы можем использовать химическое уравнение реакции сгорания алюминия в кислороде:

Calculating the Mass of Aluminum Oxide

To calculate the mass of aluminum oxide that can be obtained from the complete combustion of 81g of aluminum in oxygen, we need to consider the chemical equation for the reaction.

The balanced chemical equation for the combustion of aluminum in oxygen is:

2Al + (3/2)O2 → Al2O3

From the balanced chemical equation, we can see that 2 moles of aluminum react with (3/2) moles of oxygen to produce 1 mole of aluminum oxide.

Using the molar mass of aluminum (Al) and aluminum oxide (Al2O3), we can calculate the mass of aluminum oxide produced.

The molar mass of aluminum (Al) is approximately 27 g/mol and the molar mass of aluminum oxide (Al2O3) is approximately 102 g/mol.

Calculation

Given: Mass of aluminum (Al) = 81g

1. Calculate the number of moles of aluminum: Number of moles = Mass / Molar mass Number of moles of Al = 81g / 27 g/mol = 3 moles

2. Using the stoichiometry of the balanced chemical equation, we find the number of moles of aluminum oxide produced: According to the balanced chemical equation, 2 moles of Al produce 1 mole of Al2O3. Therefore, 3 moles of Al will produce (3/2) moles of Al2O3.

3. Calculate the mass of aluminum oxide produced: Mass of Al2O3 = Number of moles * Molar mass Mass of Al2O3 = (3/2) * 102 g/mol = 153 g

Answer

The mass of aluminum oxide that can be obtained from the complete combustion of 81g of aluminum in oxygen is 153g.