Answer:
2.83848 grams of [tex]H_{2}[/tex] are produced
Explanation:
Assuming that the reaccions continues until the total consumption of the reagents, it is possible to establish:
1 mol Al = 1 mol [tex]AlCl_{3}[/tex]
2 mol [tex]AlCl_{3}[/tex] = 3 mol [tex]H_{2}[/tex]
Then, by looking at the periodic table, it is possible to know the molar wheigt of the compounds:
1 mol Al = 29.9815 gr
1 mol [tex]H_{2}[/tex] = 2 gr
1 mol [tex]AlCl_{3}[/tex] = 136.3405 gr
Reeplacing the moles with the respectly wheights of compounds:
29.9815 gr Al = 136.3405 gr [tex]AlCl_{3}[/tex]
X gr Al = 129 gr [tex]AlCl_{3}[/tex]
So, 129 grams of [tex]AlCl_{3}[/tex] equals to 28.3673 grams of Al.
Then, following the same line of thought:
29.963 gr Al (2 moles) = 6 gr [tex]H_{2}[/tex] (3 moles)
28.3673 gr Al = Y gr [tex]H_{2}[/tex]
So, 28.3673 grams of Al equals to 2.8384 grams of [tex]H_{2}[/tex].
