So, to find the amount of AlCl3 produced, we could write the following:
[tex]4.9\text{gMgCl}2\cdot\frac{1\text{molMgCl}2}{95.2\text{gMgCl}2}\cdot\frac{2\text{molesAlCl3}}{3\text{molesMgCl}2}\cdot\frac{133.34gAlCl3}{1\text{molAlCl}3}=4.57539gAlCl3[/tex]Therefore, there are 4.57539 grams of AlCl3 produced.
