To determine how many grams of aluminum chloride ([tex]\(AlCl_3\)[/tex]) are formed when 0.25 moles of chlorine ([tex]\(Cl_2\)[/tex]) react, we need to follow these steps:
1. Analyze the Stoichiometry of the Chemical Reaction:
The given chemical equation is:
[tex]\[
2Al + 3Cl_2 \rightarrow 2AlCl_3
\][/tex]
From this equation, we can see the molar ratio between [tex]\(Cl_2\)[/tex] and [tex]\(AlCl_3\)[/tex]:
- 3 moles of [tex]\(Cl_2\)[/tex] produce 2 moles of [tex]\(AlCl_3\)[/tex].
2. Convert Moles of [tex]\(Cl_2\)[/tex] to Moles of [tex]\(AlCl_3\)[/tex]:
Since 3 moles of [tex]\(Cl_2\)[/tex] produce 2 moles of [tex]\(AlCl_3\)[/tex], we have the following ratio:
[tex]\[
\frac{2 \text{ moles of } AlCl_3}{3 \text{ moles of } Cl_2} = \frac{2}{3}
\][/tex]
Therefore, for every 1 mole of [tex]\(Cl_2\)[/tex], [tex]\(\frac{2}{3}\)[/tex] moles of [tex]\(AlCl_3\)[/tex] are produced.
Given 0.25 moles of [tex]\(Cl_2\)[/tex]:
[tex]\[
\text{Moles of } AlCl_3 = 0.25 \text{ moles of } Cl_2 \times \frac{2}{3} = 0.16666666666666666 \text{ moles of } AlCl_3
\][/tex]
3. Calculate the Mass of [tex]\(AlCl_3\)[/tex]:
The molar mass of [tex]\(AlCl_3\)[/tex] is given as 133.33 g/mol.
Using the number of moles of [tex]\(AlCl_3\)[/tex] calculated above:
[tex]\[
\text{Mass of } AlCl_3 = \text{Moles of } AlCl_3 \times \text{Molar Mass of } AlCl_3
\][/tex]
[tex]\[
\text{Mass of } AlCl_3 = 0.16666666666666666 \text{ moles} \times 133.33 \text{ g/mol} = 22.221666666666668 \text{ grams}
\][/tex]
Therefore, when 0.25 moles of [tex]\(Cl_2\)[/tex] react, 22.221666666666668 grams of aluminum chloride ([tex]\(AlCl_3\)[/tex]) are formed.
