Sure, let's solve this question step by step.
1. Determine the moles of aluminum (Al):
- Given mass of aluminum [tex]\( \text{(mass of Al)} \)[/tex]: 22.0 grams
- Molar mass of aluminum [tex]\( \text{(molar mass of Al)} \)[/tex]: 26.98 grams per mole (g/mol)
The number of moles of aluminum can be calculated using the formula:
[tex]\[
\text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}}
\][/tex]
Plugging in the values, we get:
[tex]\[
\text{moles of Al} = \frac{22.0 \text{ g}}{26.98 \text{ g/mol}} \approx 0.815 \text{ moles}
\][/tex]
2. Use stoichiometry of the balanced chemical equation:
- The balanced chemical equation is:
[tex]\[
2 \text{ Al} (s) + 3 \text{ Cl}_2 (g) \rightarrow 2 \text{ AlCl}_3 (s)
\][/tex]
- According to the equation, 2 moles of Al produce 2 moles of AlCl₃.
So, the moles of Al are directly converted to moles of AlCl₃ in a 1:1 ratio:
[tex]\[
\text{moles of AlCl}_3 = \text{moles of Al}
\][/tex]
Since we have [tex]\( 0.815 \)[/tex] moles of Al, this would mean:
[tex]\[
\text{moles of AlCl}_3 = 0.815 \text{ moles}
\][/tex]
Thus, the number of moles of aluminum chloride (AlCl₃) that can be produced from 22.0 grams of aluminum, when chlorine is in excess, is 0.815 moles.
