Answer :
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:-
[tex]2Al_{(s)}+6HBr_{(aq)}\rightarrow 2AlBr_3_{(aq)}+3H_2_{(g)}[/tex]
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
Moles of hydrogen formed = 3.5 moles
Answer:
3.5 mol H2, HBr (limiting reactant)
Explanation:
4.0 mol Al × 3 mol H2/ 2 mol Al = 6.0 mol H2
7.0 mol HB ×3 mol H2/ 6mol HBr = 3.5 mol H2
Since 7.0mol of HBr will produce less H2 than 4.0mol of Al, HBr will be the limiting reactant, and the reaction will produce 3.5mol of H2.