Question

HELPP! pls

Consider the following reaction: _2_ Al + _6_ HBr → _2_ AlBr3 + _3_ H2
When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed?
What is the limiting reactant?
What is the excess reactant?

Answer 1

"2.48 mole" of H₂ are formed. A further explanation is provided below.

Step-by-step 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:

⇒
`(Mole \ of \ H_2)/(3) =(Mole \ of HBr)/(6)`

or,

⇒
`Mole \ of \ H_2=(1)/(2)* Mole \ of \ HBr`

⇒
`=(1)/(2)* 4.96`

⇒
`=2.48 \ mole`

(b)

The limiting reactant is:

=
`HBr`

(c)

The excess reactant is:

=
`Al`