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Nijel · Answer 1 · 2023-12-23T05:05:18+0000

So, to find the amount of AlCl3 produced, we could write the following:

$4.9\text{gMgCl}2\cdot\frac{1\text{molMgCl}2}{95.2\text{gMgCl}2}\cdot\frac{2\text{molesAlCl3}}{3\text{molesMgCl}2}\cdot\frac{133.34gAlCl3}{1\text{molAlCl}3}=4.57539gAlCl3$

Therefore, there are 4.57539 grams of AlCl3 produced.

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