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:

=

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.