­­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

what's the

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.