Answer:
Moles of hydrogen formed = 3.5 moles
Explanation:
Given that:-
Moles of aluminium= 4.0 mol
Moles of hydrogen bromide = 7.0 mol
According to the reaction:-
![2Al_{(s)}+6HBr_{(aq)}\rightarrow 2AlBr_3_{(aq)}+3H_2_{(g)}](https://tex.z-dn.net/?f=2Al_%7B%28s%29%7D%2B6HBr_%7B%28aq%29%7D%5Crightarrow%202AlBr_3_%7B%28aq%29%7D%2B3H_2_%7B%28g%29%7D)
2 moles of aluminum react with 6 moles of hydrogen bromide
1 mole of aluminum react with 6/2 moles of hydrogen bromide
4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide
Moles of hydrogen bromide = 12 moles
Available moles of hydrogen bromide = 7.0 moles
Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)
The formation of the product is governed by the limiting reagent. So,
6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen
1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen
7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen
<u>Moles of hydrogen formed = 3.5 moles</u>