Stoichiometry and its Important Numericals Class 11th Chemistry
Stoichiometry: A Simplified Guide
What is Stoichiometry?
In chemistry, stoichiometry is the calculation of reactants and products in chemical reactions. It involves the quantitative relationship between reactants and products in a balanced chemical equation, expressed in terms of moles, mass, or volume.
The Concept of Stoichiometry
The term “stoichiometry” is derived from the Greek words “stoicheion” (element) and “metron” (measure). Stoichiometry provides the numerical relationship between chemical quantities in a balanced equation.
Stoichiometric Calculations
Stoichiometric calculations help us determine the amount of reactants needed to produce a specific amount of product or vice versa. Consider the following reaction:
Here, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. This gives us a simple 1:1:1 relationship in the equation.
Mole-Mole Relationship
Consider the combustion of methane:
Here, 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
Mass-Mass Relationship
The mass of reactants and products can be calculated using their molar masses. For the methane combustion example:
- Methane: 16 g/mol
- Oxygen: 32 g/mol
- Carbon dioxide: 44 g/mol
- Water: 18 g/mol
Thus, the combustion of 16 g of methane requires 64 g of oxygen, producing 44 g of carbon dioxide and 36 g of water.
Example: Calculate the volume of CO2 produced from 50 g of calcium carbonate (CaCO3) when heated.
The balanced equation is:
From the equation, 1 mole of CaCO3 produces 1 mole of CO2. Molar mass of CaCO3 is 100 g/mol. Therefore, heating 50 g of CaCO3 produces 0.5 moles of CO2, which occupies 11.35 liters at STP.
Volume-Volume Relationship
Under standard conditions (0°C and 1 atm), 1 mole of any gas occupies a volume of 22.4 liters. Using this, we can relate the volumes of gaseous reactants and products in a reaction.
Example: How much volume of chlorine gas is required to produce 11.2 L of HCl at STP?
The balanced equation is:
From the equation, 1 mole of Cl2 produces 2 moles of HCl. To produce 11.2 L of HCl, we need half the volume of Cl2, i.e., 5.6 L.
Percentage Composition
Percentage composition represents the percentage by mass of each element in a compound. For example, in magnesium carbonate (MgCO3):
In 100 g of MgCO3, we can calculate the mass percentage of each element (Mg, C, O) based on their atomic masses.
Example: Calculate the percentage composition of magnesium in MgCO3.
Molar mass of MgCO3 is 84 g/mol. Magnesium constitutes 24 g, so the percentage of Mg in MgCO3 is: