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, what is the limiting reactant?

Answer 1

Hint: 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:
$2{H_2} + {O_2} \to 2{H_2}O$
The ratio in which Hydrogen and Oxygen react is $2:1$ . 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 $moles = \dfrac{{mass(g)}}{{Molar{\text{ }}Mass(g/mol)}}$
Moles of Hydrogen ${n_{{H_2}}} = \dfrac{{10}}{{2.0159}} = 4.961mol$of Hydrogen
Moles of Oxygen ${n_{{O_2}}} = \dfrac{{15}}{{32}} = 0.4688mol$of 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 $2:1$ 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 $2moles{\text{ }}{H_2} = 1mol{\text{ }}{O_2}$
Therefore, $4.961mol{\text{ }}{H_2} = \dfrac{{4.961 \times 1}}{2}mol{\text{ }}{O_2} = 2.4805mol$ 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.
$1mol{\text{ }}{O_2} = 2{\text{ }}moles{\text{ }}{H_2}$
$0.4688mol{\text{ }}{{\text{O}}_2} = \dfrac{{0.4688 \times 2}}{1}mol{\text{ }}{H_2} = 0.9376mol$ 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.