# How many grams of aluminum are contained in 100.0 g of aluminum chloride, AlCl_3?

Apr 10, 2017

0.7500 mols of Al

#### Explanation:

100.0 g $A l C {l}_{3} \frac{1 m o l A l C {l}_{3}}{133.34 g} = .74996 m o l A l C {l}_{3}$
There is 1 mol of Aluminum per one mole of $A l C {l}_{3}$
Therefore there is .74996 mols of Al
Rounded to correct sig figs (4) that would be: .7500 mols of Al