Question

Количество вещества хлорида алюминия получившегося из 10,2 грамм оксида алюминия при взаимодействий с соляной кислотой?

Answer 1

Запишем уравнение: Al₂O₃+HCl=2AlCl₃+6H₂O
Дано: m(Al₂O₃)= 10,2 г.
Найдём молярную массу оксида алюминия: M(Al₂O₃)=2*27+3*16=54+48=102 г/моль
n(Al₂O₃) = 10,2 г/ 102 г/моль=0,1моль
n(AlCl₃)=2n(Al₂O₃)
n(AlCl₃)= 2*0,1 моль=0,2 моль
Ответ n(AlCl₃)=0,2 моль

Answer 2

Calculation of the Amount of Aluminum Chloride Produced

To calculate the amount of aluminum chloride produced from the reaction between aluminum oxide and hydrochloric acid, we need to use the balanced chemical equation for the reaction:

2 Al2O3 + 6 HCl → 2 AlCl3 + 3 H2O

From the equation, we can see that 2 moles of aluminum oxide react with 6 moles of hydrochloric acid to produce 2 moles of aluminum chloride.

Given that the mass of aluminum oxide is 10.2 grams, we need to convert this mass to moles using the molar mass of aluminum oxide.

The molar mass of aluminum oxide (Al2O3) is calculated as follows: 2(Al) + 3(O) = 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol

Now, we can calculate the number of moles of aluminum oxide: moles = mass / molar mass = 10.2 g / 101.96 g/mol = 0.100 moles

According to the balanced equation, 2 moles of aluminum oxide react to produce 2 moles of aluminum chloride. Therefore, the number of moles of aluminum chloride produced is also 0.100 moles.

To convert the number of moles of aluminum chloride to grams, we need to use the molar mass of aluminum chloride.

The molar mass of aluminum chloride (AlCl3) is calculated as follows: Al + 3(Cl) = 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol

Now, we can calculate the mass of aluminum chloride: mass = moles * molar mass = 0.100 moles * 133.33 g/mol = 13.33 grams

Therefore, the amount of aluminum chloride produced from the reaction between 10.2 grams of aluminum oxide and hydrochloric acid is 13.33 grams.

Answer 3

Calculation of the Amount of Aluminum Chloride Produced

To calculate the amount of aluminum chloride produced from the reaction between aluminum oxide and hydrochloric acid, we need to use the balanced chemical equation for the reaction:

2Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

From the equation, we can see that 2 moles of aluminum oxide react with 6 moles of hydrochloric acid to produce 2 moles of aluminum chloride.

To find the amount of aluminum chloride produced from 10.2 grams of aluminum oxide, we need to convert the mass of aluminum oxide to moles using its molar mass. The molar mass of aluminum oxide (Al₂O₃) is 101.96 g/mol.

Let's calculate the number of moles of aluminum oxide: 10.2 g Al₂O₃ × (1 mol Al₂O₃ / 101.96 g Al₂O₃) = 0.1 mol Al₂O₃

According to the balanced equation, 2 moles of aluminum oxide react to produce 2 moles of aluminum chloride. Therefore, the number of moles of aluminum chloride produced is also 0.1 mol.

Since the molar mass of aluminum chloride (AlCl₃) is 133.34 g/mol, we can calculate the mass of aluminum chloride produced: 0.1 mol AlCl₃ × (133.34 g AlCl₃ / 1 mol AlCl₃) = 13.34 g AlCl₃

Therefore, the amount of aluminum chloride produced from 10.2 grams of aluminum oxide is 13.34 grams.