Answer:
"2.48 mole" of H₂ are formed. A further explanation is provided below.
Explanation:
The given values are:
Mole of Al,
= 3.22 mole
Mole of HBr,
= 4.96 mole
Now,
(a)
The number of mole of H₂ are:
⇒ [tex]\frac{Mole \ of \ H_2}{3} =\frac{Mole \ of HBr}{6}[/tex]
or,
⇒ [tex]Mole \ of \ H_2=\frac{1}{2}\times Mole \ of \ HBr[/tex]
⇒ [tex]=\frac{1}{2}\times 4.96[/tex]
⇒ [tex]=2.48 \ mole[/tex]
(b)
The limiting reactant is:
= [tex]HBr[/tex]
(c)
The excess reactant is:
= [tex]Al[/tex]
