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?

"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:

⇒  

or,

⇒  

⇒                      

⇒                      

(b)

The limiting reactant is:

=

(c)

The excess reactant is:

=

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.