­­2K + 2HBr → 2 KBr + H2

When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)

●a). What is the limiting reactant?
●b.)What is the excess reactant?
●C.)How much product is produced?

Explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:   39.10    80.41                2.016  

            2K  +  2HBr ⟶ 2KBr + H₂

m/g:     5.5      4.04

a) Limiting reactant

(i) Calculate the moles of each reactant  

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:  

The molar ratio of H₂:K is 1:2.

From HBr:  

The molar ratio of H₂:HBr is 3:2.  

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

just did the assignment and the answer was #1 +2 n