Question

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

Answer 1

0.7500 mols of Al

Explanation:

100.0 g `AlCl_3 (1mol AlCl_3)/(133.34 g)= .74996 mol AlCl_3`
There is 1 mol of Aluminum per one mole of `AlCl_3`
Therefore there is .74996 mols of Al
Rounded to correct sig figs (4) that would be: .7500 mols of Al