Stoichiometry

STOICHEMIST

Life is all about balance. Balancing the positive and negatives halves trying to attain a neutral situation is what any system earns for. In the same way a chemical reaction has to be balanced in the right order for efficient functioning. Thus, we need to have an understanding about the how much reactants can be mixed to get the desired amount of products. This is through Stoichiometry.

What is Stoichiometry?

Stoichiometry is the way how we calculate the amount of the reactants and products in a chemical reaction by applying the rule of ratios in accordance with the law of conservation of mass.

As the law of conservation of mass states, the amount of reactants will be equal to the amount of products. This correlation is used to calculate the amount of one of the reactants such that the complete amount of the other reactant can be used up in the reaction avoiding any wastage of resources. If one of the reactants is not in the required quantity to utilize the amount of the other, that reactant is called as the limiting reagent of the reaction. Hence, if we have the quantity of one reactant of a balanced chemical reaction then we can calculate the amount of the other reactant with the help of the Stoichiometry.

Let us try and understand it better with the help of an example.

Example:

\(CH_4 + 2O_2 \rightarrow CO_2 +2H_2O\)

In the simple above reaction, we have methane reacting with oxygen to give out carbon dioxide and water releasing heat in the process as well. Being termed as a complete reaction, for proper and efficient combustion, we need 1 mole of methane with 2 moles of carbon dioxide giving out 1 mole of carbon dioxide with 2 moles of water.

If we have a given weight of methane, then we can calculate the amount of oxygen required in weight to obtain an efficient combustion reaction owing to the empirically determined ratio of the products.

Suppose we have a mass of 8 gms of methane, let us calculate the amount of oxygen we would require for complete combustion.

Step 1: Find the number of moles of the reactant

To calculate the number of moles in 8 gms of methane,

Number of moles = \(\frac{{8 \; g \; CH_4}\times{1 mol CH_4}}{16.04 \; g}\)

= 0.49875 moles

Step 2: Ratio of the number of moles of the other reactant

Mole ratio between oxygen and methane = \(\frac{2}{1}=2\)

Thus number of moles of oxygen = 2 x 0.49875 = 0.9975 mol

Step 3: Finding the mass from number of moles

The molecular mass of oxygen = 15.9994 g/mol

Thus the weight of oxygen = 0.9975 mol x 15.9994 g/mol = 15.9594015 gms

Thus for complete combustion, we would require 15.9594015 gms of oxygen to obtain utilize the reactants completely.

Similarly, we can use stoichiometry to completely use all of the reactants and obtain the maximum amount of the product with minimal wastage of substances.

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