Answer:
≈ 2.74 mol [tex]AlCl_{3}[/tex]
(I rounded to 3 sig figs because the starting amount has 3 sig figs.)
Explanation:
[tex]\frac{365g AlCl_{3}}{1}*\frac{1 mol AlCl_{3}}{133.34g AlCl_{3}} =[/tex] 2.73736313184 mol [tex]AlCl_{3}[/tex] ≈ 2.74 mol [tex]AlCl_{3}[/tex]
Hope this helps!
