Answer:
2.83848 grams of
are produced
Step-by-step explanation:
Assuming that the reaccions continues until the total consumption of the reagents, it is possible to establish:
1 mol Al = 1 mol
![AlCl_(3)](https://img.qammunity.org/2020/formulas/chemistry/high-school/geh02sqgl7r5g2899lyp0g563vmpywj5jo.png)
2 mol
= 3 mol
![H_(2)](https://img.qammunity.org/2020/formulas/chemistry/middle-school/qj3jlkx2g8ca3kmhyhel7me8a593mr9yed.png)
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
= 2 gr
1 mol
= 136.3405 gr
Reeplacing the moles with the respectly wheights of compounds:
29.9815 gr Al = 136.3405 gr
![AlCl_(3)](https://img.qammunity.org/2020/formulas/chemistry/high-school/geh02sqgl7r5g2899lyp0g563vmpywj5jo.png)
X gr Al = 129 gr
![AlCl_(3)](https://img.qammunity.org/2020/formulas/chemistry/high-school/geh02sqgl7r5g2899lyp0g563vmpywj5jo.png)
So, 129 grams of
equals to 28.3673 grams of Al.
Then, following the same line of thought:
29.963 gr Al (2 moles) = 6 gr
(3 moles)
28.3673 gr Al = Y gr
![H_(2)](https://img.qammunity.org/2020/formulas/chemistry/middle-school/qj3jlkx2g8ca3kmhyhel7me8a593mr9yed.png)
So, 28.3673 grams of Al equals to 2.8384 grams of
.