[tex]\[
2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3
\][/tex]

How many grams of aluminum are required to produce 8.00 moles of aluminum chloride?

Molar mass of Al: [tex]\(26.98 \, \text{g/mol}\)[/tex]

[tex]\[
[?] \, \text{g Al}
\][/tex]



Answer :

To determine how many grams of aluminum are required to produce 8.00 moles of aluminum chloride (AlCl₃), we can follow these steps:

1. Determine the molar relationship from the balanced chemical equation:
The balanced chemical equation given is:
[tex]\[ 2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3 \][/tex]
From the equation, we see that 2 moles of aluminum (Al) produce 2 moles of aluminum chloride (AlCl₃).

2. Identify the molar relationship:
The stoichiometry tells us that 1 mole of aluminum (Al) produces 1 mole of aluminum chloride (AlCl₃).

3. Calculate the moles of aluminum needed:
Given that we need to produce 8.00 moles of aluminum chloride (AlCl₃), the moles of aluminum required are equal to the moles of AlCl₃ produced. Therefore, we need 8.00 moles of aluminum (Al).

4. Use the molar mass of aluminum to find the required grams:
The molar mass of aluminum (Al) is given as 26.98 g/mol.

5. Calculate the mass of aluminum:
To convert moles of aluminum to grams, use the formula:
[tex]\[ \text{mass} = \text{moles} \times \text{molar mass} \][/tex]
Substituting in the values:
[tex]\[ \text{mass of Al} = 8.00 \text{ moles} \times 26.98 \text{ g/mol} = 215.84 \text{ grams} \][/tex]

Thus, to produce 8.00 moles of aluminum chloride (AlCl₃), you need 215.84 grams of aluminum (Al).