Question

How many moles are equivalent to 365 g of aluminum chloride? Using the dimensions analysis. Show work

Answer 1

Answer:

≈ 2.74 mol $AlCl_{3}$

(I rounded to 3 sig figs because the starting amount has 3 sig figs.)

Explanation:

$\frac{365g AlCl_{3}}{1}*\frac{1 mol AlCl_{3}}{133.34g AlCl_{3}} =$ 2.73736313184 mol $AlCl_{3}$ ≈ 2.74 mol $AlCl_{3}$

Hope this helps!