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Question: Given the equation: \(2{H_2} + {O_2} \to 2{H_2}O\), if given 10g \({H_2}\) gas and 15g \({O_2}\) gas...

Given the equation: 2H2+O2→2H2O2{H_2} + {O_2} \to 2{H_2}O, if given 10g H2{H_2} gas and 15g O2{O_2} gas, what is the limiting reactant?

Explanation

Solution

The limiting reagent is the compound that is fully consumed in a chemical reaction. It is also known as Limiting reactant or limiting reagent. The product formed depends on the limiting reagent, since without it the reaction cannot continue.

Complete answer:
The balanced chemical reaction can be given as:
2H2+O2→2H2O2{H_2} + {O_2} \to 2{H_2}O
The ratio in which Hydrogen and Oxygen react is 2:12:1 . Which means that 2 moles of Hydrogen will need 1 mol of dioxygen. This means that, regardless of the moles of Oxygen we’ll always need double the amount of Hydrogen.
First let us find the no. of moles of each given to us. We will use the formula moles=mass(g)Molar Mass(g/mol)moles = \dfrac{{mass(g)}}{{Molar{\text{ }}Mass(g/mol)}}
Moles of Hydrogen nH2=102.0159=4.961mol{n_{{H_2}}} = \dfrac{{10}}{{2.0159}} = 4.961molof Hydrogen
Moles of Oxygen nO2=1532=0.4688mol{n_{{O_2}}} = \dfrac{{15}}{{32}} = 0.4688molof Dioxygen
From no. of moles, we can see that the amount of Hydrogen is massively excessive. The amount of Hydrogen is way more than the required 2:12:1 ratio.
Now, to find the limiting reagent let us know the no. of moles of Oxygen required to react with 4.961 moles of Hydrogen. We know that 2moles H2=1mol O22moles{\text{ }}{H_2} = 1mol{\text{ }}{O_2}
Therefore, 4.961mol H2=4.961×12mol O2=2.4805mol4.961mol{\text{ }}{H_2} = \dfrac{{4.961 \times 1}}{2}mol{\text{ }}{O_2} = 2.4805mol of Oxygen is required.
But, we have only 0.4988 moles of Oxygen. Hence Oxygen is the limiting reagent.
Now, let us find the amount of excess hydrogen present in the reaction mixture. For that we’ll find the no. of moles of Hydrogen required to react with 0.4688 moles of dioxygen.
1mol O2=2 moles H21mol{\text{ }}{O_2} = 2{\text{ }}moles{\text{ }}{H_2}
0.4688mol O2=0.4688×21mol H2=0.9376mol0.4688mol{\text{ }}{{\text{O}}_2} = \dfrac{{0.4688 \times 2}}{1}mol{\text{ }}{H_2} = 0.9376mol of hydrogen
Apart from this 0.9376 mol of hydrogen, rest all is in excess.

Note:
The two methods followed to find the limiting agent are:
i) By comparing the amount of reactants. This is useful in the reaction where two reactants are present. In this the balanced equation is used to determine the amount of another reactant, by considering the first. If the second is more than the first, the first is the limiting reagent.
ii) By comparing the amount of product formed. The amount of product formed from the reactant present, is determined by the chemical equation. The reactant that forms the smallest amount of product is the limiting reagent.