Question

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

Answer 1

O2

Step-by-step explanation:

2020 edge

Answer 2

Step-by-step 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.