What is stoichiometry? and it's calculations
What is stoichiometry? and it's calculations
Dear student
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:
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 =
= 0.49875 moles
Step 2: Ratio of the number of moles of the other reactant
Mole ratio between oxygen and methane =
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.
Hope you understand it
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:
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 =
= 0.49875 moles
Step 2: Ratio of the number of moles of the other reactant
Mole ratio between oxygen and methane =
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.
Hope you understand it
