Given The Chemical Reaction:$\[ 2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3 \\]How Many Grams Of Aluminum Chloride, \[$\text{AlCl}_3\$\], Are Formed When \[$0.25 \, \text{mol Cl}_2\$\] React?Molar Mass Of

Understanding the Chemical Reaction

The given chemical reaction is: $2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3$. This reaction involves the combination of aluminum (Al) and chlorine gas (Cl$_2$) to form aluminum chloride (AlCl$_3$). The reaction is a classic example of a synthesis reaction, where two or more substances combine to form a new compound.

Stoichiometry and Molar Ratios

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. In this reaction, the molar ratio between Cl$_2$ and AlCl$_3$ is 3:2. This means that for every 3 moles of Cl$_2$ that react, 2 moles of AlCl$_3$ are formed.

Calculating the Number of Moles of Aluminum Chloride Formed

To calculate the number of moles of AlCl$_3$ formed, we need to use the molar ratio between Cl$_2$ and AlCl$_3$. Given that 0.25 mol of Cl$_2$ react, we can calculate the number of moles of AlCl$_3$ formed as follows:

$$\text{moles of AlCl}_3 = \frac{\text{moles of Cl}_2 \times \text{molar ratio of AlCl}_3}{\text{molar ratio of Cl}_2}$$

$$\text{moles of AlCl}_3 = \frac{0.25 \, \text{mol} \times 2}{3}$$

$$\text{moles of AlCl}_3 = \frac{0.5}{3}$$

$$\text{moles of AlCl}_3 = 0.1667 \, \text{mol}$$

Calculating the Mass of Aluminum Chloride Formed

To calculate the mass of AlCl$_3$ formed, we need to use the molar mass of AlCl$_3$. The molar mass of AlCl$_3$ is 133.34 g/mol. We can calculate the mass of AlCl$_3$ formed as follows:

$$\text{mass of AlCl}_3 = \text{moles of AlCl}_3 \times \text{molar mass of AlCl}_3$$

$$\text{mass of AlCl}_3 = 0.1667 \, \text{mol} \times 133.34 \, \text{g/mol}$$

$$\text{mass of AlCl}_3 = 22.21 \, \text{g}$$

Conclusion

In conclusion, when 0.25 mol of Cl$_2$ react, 0.1667 mol of AlCl$_3$ are formed. The mass of AlCl$_3$ formed is 22.21 g.

Molar Mass of Aluminum Chloride

The molar mass of AlCl$_3$ is 133.34 g/mol. This value is calculated by adding the atomic masses of aluminum (Al) and three chlorine atoms (Cl).

Atomic Masses of Elements

The atomic masses of elements are:

• Aluminum (Al): 26.98 g/mol
• Chlorine (Cl): 35.45 g/mol

Calculating the Molar Mass of Aluminum Chloride

To calculate the molar mass of AlCl$_3$, we add the atomic masses of Al and three Cl atoms:

$$\text{molar mass of AlCl}_3 = \text{atomic mass of Al} + 3 \times \text{atomic mass of Cl}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 3 \times 35.45 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 106.35 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 133.33 \, \text{g/mol}$$

Molar Mass of Aluminum Chloride (AlCl$_3$)

The molar mass of AlCl$_3$ is 133.34 g/mol.

Atomic Masses of Elements

The atomic masses of elements are:

• Aluminum (Al): 26.98 g/mol
• Chlorine (Cl): 35.45 g/mol

Calculating the Molar Mass of Aluminum Chloride

To calculate the molar mass of AlCl$_3$, we add the atomic masses of Al and three Cl atoms:

$$\text{molar mass of AlCl}_3 = \text{atomic mass of Al} + 3 \times \text{atomic mass of Cl}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 3 \times 35.45 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 106.35 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 133.33 \, \text{g/mol}$$

Molar Mass of Aluminum Chloride (AlCl$_3$)

The molar mass of AlCl$_3$ is 133.34 g/mol.

Atomic Masses of Elements

The atomic masses of elements are:

• Aluminum (Al): 26.98 g/mol
• Chlorine (Cl): 35.45 g/mol

Calculating the Molar Mass of Aluminum Chloride

To calculate the molar mass of AlCl$_3$, we add the atomic masses of Al and three Cl atoms:

$$\text{molar mass of AlCl}_3 = \text{atomic mass of Al} + 3 \times \text{atomic mass of Cl}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 3 \times 35.45 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 106.35 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 133.33 \, \text{g/mol}$$

Molar Mass of Aluminum Chloride (AlCl$_3$)

The molar mass of AlCl$_3$ is 133.34 g/mol.

Atomic Masses of Elements

The atomic masses of elements are:

• Aluminum (Al): 26.98 g/mol
• Chlorine (Cl): 35.45 g/mol

Calculating the Molar Mass of Aluminum Chloride

To calculate the molar mass of AlCl$_3$, we add the atomic masses of Al and three Cl atoms:

$$\text{molar mass of AlCl}_3 = \text{atomic mass of Al} + 3 \times \text{atomic mass of Cl}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 3 \times 35.45 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 106.35 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 133.33 \, \text{g/mol}$$

Molar Mass of Aluminum Chloride (AlCl$_3$)

The molar mass of AlCl$_3$ is 133.34 g/mol.

Atomic Masses of Elements

The atomic masses of elements are:

• Aluminum (Al): 26.98 g/mol
• Chlorine (Cl): 35.45 g/mol

Calculating the Molar Mass of Aluminum Chloride

To calculate the molar mass of AlCl$_3$, we add the atomic masses of Al and three Cl atoms:

$$\text{molar mass of AlCl}_3 = \text{atomic mass of Al} + 3 \times \text{atomic mass of Cl}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 3 \times 35.45 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 106.35 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 133.33 \, \text{g/mol}$$

Molar Mass of Aluminum Chloride (AlCl$_3$)

The molar mass of AlCl$_3$ is 133.34 g/mol.

Atomic Masses of Elements

The atomic masses of elements are:

• Aluminum (Al): 26.98 g/mol
• Chlorine (Cl): 35.45 g/mol

Calculating the Molar Mass of Aluminum Chloride

To calculate the molar mass of AlCl$_3$, we

Q: What is the chemical reaction for the formation of aluminum chloride?

A: The chemical reaction for the formation of aluminum chloride is: $2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3$. This reaction involves the combination of aluminum (Al) and chlorine gas (Cl$_2$) to form aluminum chloride (AlCl$_3$).

Q: What is the molar ratio between Cl$_2$ and AlCl$_3$ in the reaction?

A: The molar ratio between Cl$_2$ and AlCl$_3$ in the reaction is 3:2. This means that for every 3 moles of Cl$_2$ that react, 2 moles of AlCl$_3$ are formed.

Q: How many moles of AlCl$_3$ are formed when 0.25 mol of Cl$_2$ react?

A: To calculate the number of moles of AlCl$_3$ formed, we use the molar ratio between Cl$_2$ and AlCl$_3$. Given that 0.25 mol of Cl$_2$ react, we can calculate the number of moles of AlCl$_3$ formed as follows:

$$\text{moles of AlCl}_3 = \frac{\text{moles of Cl}_2 \times \text{molar ratio of AlCl}_3}{\text{molar ratio of Cl}_2}$$

$$\text{moles of AlCl}_3 = \frac{0.25 \, \text{mol} \times 2}{3}$$

$$\text{moles of AlCl}_3 = \frac{0.5}{3}$$

$$\text{moles of AlCl}_3 = 0.1667 \, \text{mol}$$

Q: What is the mass of AlCl$_3$ formed when 0.25 mol of Cl$_2$ react?

A: To calculate the mass of AlCl$_3$ formed, we use the molar mass of AlCl$_3$. The molar mass of AlCl$_3$ is 133.34 g/mol. We can calculate the mass of AlCl$_3$ formed as follows:

$$\text{mass of AlCl}_3 = \text{moles of AlCl}_3 \times \text{molar mass of AlCl}_3$$

$$\text{mass of AlCl}_3 = 0.1667 \, \text{mol} \times 133.34 \, \text{g/mol}$$

$$\text{mass of AlCl}_3 = 22.21 \, \text{g}$$

Q: What is the molar mass of aluminum chloride (AlCl$_3$)?

A: The molar mass of AlCl$_3$ is 133.34 g/mol. This value is calculated by adding the atomic masses of aluminum (Al) and three chlorine atoms (Cl).

Q: How do you calculate the molar mass of aluminum chloride (AlCl$_3$)?

A: To calculate the molar mass of AlCl$_3$, we add the atomic masses of Al and three Cl atoms:

$$\text{molar mass of AlCl}_3 = \text{atomic mass of Al} + 3 \times \text{atomic mass of Cl}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 3 \times 35.45 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 106.35 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 133.33 \, \text{g/mol}$$

Q: What are the atomic masses of aluminum (Al) and chlorine (Cl)?

A: The atomic masses of elements are:

• Aluminum (Al): 26.98 g/mol
• Chlorine (Cl): 35.45 g/mol

Q: How do you calculate the molar mass of aluminum chloride (AlCl$_3$) using the atomic masses of Al and Cl?

A: To calculate the molar mass of AlCl$_3$, we add the atomic masses of Al and three Cl atoms:

$$\text{molar mass of AlCl}_3 = \text{atomic mass of Al} + 3 \times \text{atomic mass of Cl}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 3 \times 35.45 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 26.98 \, \text{g/mol} + 106.35 \, \text{g/mol}$$

$$\text{molar mass of AlCl}_3 = 133.33 \, \text{g/mol}$$

Q: What is the molar mass of aluminum chloride (AlCl$_3$)?

A: The molar mass of AlCl$_3$ is 133.34 g/mol.

Q: How do you calculate the mass of aluminum chloride (AlCl$_3$) formed in a reaction?

A: To calculate the mass of AlCl$_3$ formed, we use the molar mass of AlCl$_3$ and the number of moles of AlCl$_3$ formed:

$$\text{mass of AlCl}_3 = \text{moles of AlCl}_3 \times \text{molar mass of AlCl}_3$$

$$\text{mass of AlCl}_3 = 0.1667 \, \text{mol} \times 133.34 \, \text{g/mol}$$

$$\text{mass of AlCl}_3 = 22.21 \, \text{g}$$

Q: What is the relationship between the number of moles of Cl$_2$ and the number of moles of AlCl$_3$ formed in the reaction?

A: The molar ratio between Cl$_2$ and AlCl$_3$ in the reaction is 3:2. This means that for every 3 moles of Cl$_2$ that react, 2 moles of AlCl$_3$ are formed.

Q: How do you calculate the number of moles of AlCl$_3$ formed when a certain number of moles of Cl$_2$ react?

A: To calculate the number of moles of AlCl$_3$ formed, we use the molar ratio between Cl$_2$ and AlCl$_3$. Given that a certain number of moles of Cl$_2$ react, we can calculate the number of moles of AlCl$_3$ formed as follows:

$$\text{moles of AlCl}_3 = \frac{\text{moles of Cl}_2 \times \text{molar ratio of AlCl}_3}{\text{molar ratio of Cl}_2}$$

$$\text{moles of AlCl}_3 = \frac{\text{moles of Cl}_2 \times 2}{3}$$

Q: What is the relationship between the mass of Cl$_2$ and the mass of AlCl$_3$ formed in the reaction?

A: The mass of AlCl$_3$ formed is directly proportional to the mass of Cl$_2$ that reacts. This is because the molar mass of AlCl$_3$ is constant, and the number of moles of AlCl$_3$ formed is directly proportional to the number of moles of Cl$_2$ that react.

Q: How do you calculate the mass of AlCl$_3$ formed when a certain mass of Cl$_2$ reacts?

A: To calculate the mass of AlCl$_3$ formed, we use the molar mass of AlCl$_3$ and the number of moles of AlCl$_3$ formed. We can calculate the number of moles of AlCl$_3$ formed using the molar ratio between Cl$_2$ and AlCl$_3$:

$$\text{moles of AlCl}_3 = \frac{\text{moles of Cl}_2 \times \text{molar ratio of AlCl}_3}{\text{molar ratio of Cl}_2}$$

$$\text{moles of AlCl}_3 = \frac{\text{moles of Cl}_2 \times 2}{3}$$

We can then calculate the mass of AlCl$_3$ formed using the molar mass of AlCl$_3$:

$$\text{mass of AlCl}_3 = \text{moles of AlCl}_3 \times \text{molar mass of AlCl}_3$$

$$\text{mass of AlCl}_3 = 0.1667 \, \text{mol} \times 133.34 \, \text{g/mol}$$

$$\text{mass of AlCl}_3 = 22.21 \, \text{g}$$