Question

Вычислить количество вещества оксида алюминия, образовавшегося при взаимодействии 81г алюминия с кислородом.

Answer 1

Ответ:

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.

Объяснение:

Answer 2

Calculating the Amount of Aluminum Oxide Formed

To calculate the amount of substance of aluminum oxide formed when 81g of aluminum reacts with oxygen, we can use the balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide:

4Al + 3O2 → 2Al2O3

Using this equation, we can calculate the amount of aluminum oxide formed.

Steps to Calculate:

1. Calculate the Amount of Aluminum (Al) in Moles: - The molar mass of aluminum (Al) is 27g/mol. - Number of moles of aluminum (n(Al)) = Mass of aluminum (m(Al)) / Molar mass of aluminum (M(Al)) = 81g / 27g/mol = 3 moles.

2. Calculate the Amount of Aluminum Oxide (Al2O3) in Moles: - From the balanced chemical equation, 4 moles of aluminum react to form 2 moles of aluminum oxide. - Therefore, the number of moles of aluminum oxide formed (n(Al2O3)) = 3 moles of Al * (2 moles of Al2O3 / 4 moles of Al) = 1.5 moles.

3. Calculate the Mass of Aluminum Oxide Formed: - The molar mass of aluminum oxide (Al2O3) is 27*2 + 16*3 = 54 + 48 = 102g/mol. - Mass of aluminum oxide formed (m(Al2O3)) = Number of moles of aluminum oxide (n(Al2O3)) * Molar mass of aluminum oxide (M(Al2O3)) = 1.5 moles * 102g/mol = 153g.

Answer:

The amount of substance of aluminum oxide formed when 81g of aluminum reacts with oxygen is 153g.