The following equation shows the formation of water (h2o, 18.02 g/mol) from hydrogen (h2, 2.02 g/mol) and oxygen (o2, 32.00 g/mol): 2h2 + o2 2h2o if 3.5 g h2 react with 18.7 g o2, what is the limiting reactant?
a> n> s: o2

Explanation:

Data Given:

Amount of hydrogen (H₂) = 3.5 g

Oxygen gas (O₂) = 18.7 g

molar mass of H₂O = 18.02 g/mol

molar mass of O₂ = 32 g/mol

molar mass of H₂ = 2.02 g/mol

limiting reactant = ?

Solution:

limiting reactant: The amount of reactant that consume first in a reaction is termed as limiting reagent or limiting reactant.

Excess reagent: the amount of reactant that remain un-reacted at the end of the reaction is termed as excess reagent.

Now to look for the limiting regent we have to look at reaction

Reaction:

        2H₂ + O₂ > 2H₂O

Convert moles to masses

              2H₂      + O₂    >    2H₂O

              2 mol      1 mol

As we know

molar mass of O₂ = 32 g/mol

molar mass of H₂ = 2.02 g/mol

So we can write it as below

                   2H₂                +      O₂         >        2H₂O

              2 mol (2 g/mol)       1 mol (32 g/mol)

                    4 g                            32 g

To know with how much hydrogen will react with oxygen apply unity formula

                4 g of Hydrogen ≅ 32 g of O₂

                 g of Hydrogen ≅ 18.7 g of O₂

So,

                  g of H₂ = 4 g x 18.7 g / 32 g

                  g of H₂ = 2.33 g

So,

18.7 g of oxygen fully consumed and react with 2.33 g of Hydrogen and 1.16 g of hydrogen remain unreacted.

Oxygen is limiting reactant

From this balance reaction we come to know that 18.7 g of oxygen react with 2.33 g of hydrogen molecule. It means that amount of oxygen is less and it will consume first and act as limiting reactant.

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