Question

Рассчитайте массу хлорида алюминия, образующегося при взаимодействии избытка алюминия с 2,24 л (н.у.) хлора. ( С решением пожалуйста)

Answer 1

Пишем реакцию
2Al + 3Cl2 = 2AlCl3
Считаем количество хлора V/Vm = 2.24/22.4 = 0.1 моль.
За уравнением хлорида алюминия в 1,5 раза меньше, т.е. 0,1/1,5 = 0.067 моль. Считаем массу 0,067*133,5 = 8,9 г.
Ответ: 8,9 грамм.

Answer 2

Calculating the Mass of Aluminum Chloride

To calculate the mass of aluminum chloride formed when an excess of aluminum reacts with 2.24 L of chlorine gas, we can use the balanced chemical equation for the reaction and the principles of stoichiometry.

The balanced chemical equation for the reaction between aluminum and chlorine is: 2 Al + 3 Cl2 → 2 AlCl3

First, we need to find the number of moles of chlorine present in 2.24 L of chlorine gas at standard temperature and pressure (STP), which is 0°C and 1 atm pressure.

From Source we know that 1 mole of any gas at STP occupies 22.4 L. Therefore, the number of moles of chlorine gas can be calculated as follows: n = V/22.4 n = 2.24/22.4 n = 0.1 moles

Now, using the balanced chemical equation, we can determine the number of moles of aluminum chloride formed. According to Source the molar mass of aluminum chloride (AlCl3) is 133.34 g/mol.

From the balanced chemical equation, we know that 2 moles of aluminum react with 3 moles of chlorine to produce 2 moles of aluminum chloride.

Therefore, the number of moles of aluminum chloride formed will be: n(AlCl3) = (2/3) * n(Cl2) n(AlCl3) = (2/3) * 0.1 n(AlCl3) = 0.0667 moles

Finally, we can calculate the mass of aluminum chloride formed using the formula: mass = number of moles * molar mass mass = 0.0667 moles * 133.34 g/mol mass ≈ 8.89 grams

The mass of aluminum chloride formed when an excess of aluminum reacts with 2.24 L of chlorine gas is approximately 8.89 grams.