Periodic Table Analysis

A complete Guide on Introduction to Stoichiometry 

Learning stoichiometry is very vital in Chemistry but there are certain guidelines that one need to 

follow to fully grasp the concept of stoichiometry very well. 

These guidelines help not only in the understanding of the concept but also in the stoichiometry calculations. 

I will talk about six critical areas in stoichiometry that every student that want to study stoichiometry must understand. 

 The six sections I want to discuss are listed below; 

  1. Meaning of Stoichiometry 
  1. Mole concept  
  1. Balancing of Chemical equations 
  1. Solving Stoichiometry Problems 
  1. Calculating limiting and excess reagent 
  1. Calculating Theoretical yield 

Meaning of Stoichiometry  

Stoichiometry is he branch of Chemistry that deals with the quantitative relationships between the reactants and products in a balanced chemical equation.  

Stoichiometry also uses balance chemical equations to determine the amounts of reactants needed to yield a given amount of product, or the quantity of product that will be formed from a given amount of reactants.  

Solving stoichiometry Problems involving involves following steps and then practicing a lot of stoichiometry exercises.

Additionally, it includes the concept of limiting reagent (reactant that finished first in a chemical reaction) and excess reagents (reactant that remained in a chemical in a chemical reaction). 

Finally, stoichiometry covers also covers the area of theoretical yield and percentage yield of products in a chemical reaction. 

The Mole Concept and Stoichiometry  

The mole concept and stoichiometry are fundamental concepts in chemistry and the basis of stoichiometry problems and Calculations. 

Definition of a Mole? 

A mole is defined as the amount that contains as many elementary particles (such as atoms, molecules, or ions) as there are particles as there are in 12 grams of Carbon 12. 

The mole is a unit of measurement used to quantify the amount of a substance in a chemical reaction and very importantly gives us a simple way to convert from mole to mass, volume and the number of entities. 

The mole concept can be applied in solving stoichiometry problems by comparing the amount of reactants and products quantitatively.

Conversion of Mole to Mass Stoichiometry 

Mole concept is used in converting mole to mass in grams simply by using the molar mass of the substance. 

Example 1 

What mass of magnesium chloride is equal to 3.5 moles of the substance? 

Solution  

Conversion factor  

1 mole of a substance == Molar mass of the substance 

1 mole of MgCl2 ===== 95 grams (Using Periodic table) 

3.5 moles of MgCl2 === X 

X= 3.5 mole X 95 grams /1 mole 

X= 332.5 grams of Magnesium Chloride 

Example 2 

What number of moles of Calcium carbonate will produce 188.8 grams of Calcium Carbonate? 

Solution  

Conversion factor  

1 mole of a substance == Molar mass of the substance 

1 mole of CaCO3 === 100 grams (From Periodic table) 

X moles of CaCO3 ===188.8 grams 

X = 1 mole X 188.8 grams /100 grams 

X = 1.888 mole of Calcium Carbonate 

Conversion of Mole to Volume Stoichiometry  

Mole concept is used in converting mole to volume by multiplying by molar volume at standard temperature and pressure (22.4 Litres or 22.4 dm3

Example 1 

What is the volume occupied by 2.8 moles of ammonia gas at stp? 

Solution  

Conversion factor 

1 mole of a gas == Molar Volume (22.4 Litres or 22.4 dm3

1 mole of NH3 === ==22.4 dm3 

2.8 moles of NH3 === =X 

X = 62.72 dm3 

Example 2 

What mole is required to yield 78.9 litres of carbon IV oxide gas? 

Solution  

1 mole of gas = molar volume (22.4 Litres or 22.4 dm3

1 mole of CO2 === 22.4 Litres 

X moles of CO2 === 78.9 Litres 

X = 1 mole X 78.9 Litres/22.4 Litres 

X= 3.522 moles of Carbon IV oxide 

Conversion of Mole to Number of entities 

Mole can be converted to number of particles (number of entities) and this can be used as Avogadro’s number Particles. 

Example 1 

How many number of particles are there in 3.4 moles of sodium? 

Solution  

1 mole of substance = Avogadro’s number (Number of particles) 

1 mole of Na ==== 6.02 x 1023 particles  

3.4 moles of Na==== X 

X = 3.4 moles X 6.02 x 1023 Particles /1 mole 

X = 2.064 x 1024 Particles 

Example 2 

What mole of Oxygen is equal to 3.4 x 1024 Particles? 

Solution 

1 mole of gas = molar volume (22.4 Litres or 22.4 dm3

1 mole of O2 ======= 6.02 x 1023 Particles  

X moles of O2 ====== 3.4 x 1024 Particles 

X = 3.4 x 1024 Particles X 1 mole / 6.02 x 1023 

X=5.65 moles of Oxygen 

Balancing of Chemical Equations 

Balancing chemical equations is very simple if you can follow simple instructions detailed on how to balance chemical equations in chemistry. 

I will lay down some simple rules which involves  

  1. Do not write any number in between the compound 
  1. Do not write any number as subscript  
  1. Do not change the subscript 
  1. Write the number Infront as coefficients  

To help us understand the basics at least, let’s look at few examples of these equations 

Steps in Balancing Chemical Equations 

  1. Check the number of atoms on both sides of the equation 
  1. Ensure the same number of atoms on both sides of the equation 

How to balance chemical equation  

Although, I will not go step by step because I have written a post on how to balance chemical equations step by step. 

Example 1 

Let’s balance the equation on the reaction of magnesium with hydrochloric acid  

Mg + HCl ====MgCl2 + H2 

Solution  

Step 1: Check the number of atoms on both sides 

Mg + HCl ====MgCl2 + H2 

Checking the number of atoms on both sides 

1Mg 1H 1Cl ==== 1Mg 2Cl 2H 

Step 2: Ensure the same number of atoms on both sides and balance 

Mg + 2HCl ====MgCl2 + H2 

Crosscheck and verify that the equation is now balanced 

Example 2 

Balance this equation on the reaction of iron with hydrochloric acid. 

Fe + HCl ===FeCl3 + H2 

Solution  

Step 1: Check the number of atoms on both sides 

Fe + HCl ====FeCl3 + H2 

Checking the number of atoms on both sides 

1Fe 1H 1Cl ==== 1Fe 3Cl 2H 

Step 2: Ensure the same number of atoms on both sides and balance 

2Fe + 6HCl ====2FeCl3 + 3H2 

Crosscheck and verify that the equation is now balanced 

Example 3 

Show how to balance the equation of the reaction between sodium in cold water 

Na + H2O====NaOH + H2 

Solution  

Step 1: Check the number of atoms on both sides 

Na + H2O====NaOH + H2 

Checking the number of atoms on both sides 

1Na 2H 1O ==== 1Na 1O 3H 

Step 2: Ensure the same number of atoms on both sides and balance 

Na + 2H2O====2NaOH + H2 

Crosscheck and verify that the equation is now balanced 

Solving Stoichiometry Problems 

Yes, stoichiometry invariably involves solving stoichiometry calculations but no worries, oi have shared several steps and solve d problems in solving stoichiometry. 

However, let recapitulate few steps involved  

Solving stoichiometry will involves 

  1. Balance the given balanced chemical equation for the reaction. 
  1. Identify the given information and determine what quantity you’re being asked to find. This could be the mass, volume, or number of moles of a reactant or product. 
  1. Then convert the given quantity to the appropriate units, if necessary. Use the stoichiometric ratios in the balanced equation to convert the given quantity to the quantity you’re being asked to find.  

1 will solve only two examples to illustrate the stoichiometry calculation. 

Example 1 

How many moles of sodium chloride is produced by 3.4 moles of sodium with chlorine. 

2Na+ Cl2 ====2NaCl 

Solution  

2Na+ Cl2 ====2NaCl 

From the balanced equation 

2 moles of Na ==== 2 moles of NaCl 

3.4 moles of Na ==== X 

X= 3.4 x 2/ 2 = 3.4 moles of Sodium Chloride 

Example 2 

What mass of Carbon IV oxide would be evolved by the reaction of 5 moles of methane with oxygen? 

CH4 + 2O2 ====CO2 + 2H2O 

Solution  

1 mole of CH4 === 1 mole of CO2 

5 moles of CH4 === X 

X = 5 moles x 1 mole / 1 mole 

X= 5 moles of CO2 

1 mole of CO2 === 22.4 dm3 

5 moles of CO2 === X 

X= 5 x 22.4 /1 = 112 dm3 

Limiting and Excess reagent  

Limiting reagent is the reactant that finished during a chemical reaction while the excess reagent or reactant is the one that remained after the reaction.

I have a post on how to deduce the limiting and excess reagent in chemical reactions.

Calculating Theoretical yield and Percentage Yield 

Theoretical yield is the maximum amount of product that can be produced in a chemical reaction, based on the stoichiometry calculation of the reaction and assuming that all of the limiting reactant is completely consumed to form product.  

Additionally. Theoretical yield could also be defined as the amount of product that would be obtained if the reaction proceeded with 100% efficiency, without any errors in experimentation. 

Actual yield on the other hand refers to the amount of product obtained from a chemical reaction or a manufacturing process, as measured by experimentation or observation. It is typically expressed as a percentage of the theoretical yield, which is the maximum amount of product that could be obtained under ideal conditions. 

Percentage yield is defined as a measure of the efficiency of a chemical reaction, and it is expressed as actual yield over theoretical yield expressed as a percentage.  

The theoretical yield is the maximum amount of product that could be obtained under ideal conditions, while the actual yield is the amount of product that is obtained in each reaction. 

Mathematically  

Percentage yield = Actual yield/Theoretical yield X 100% 

Example  

39 grams of magnesium Chloride was Produced by the reaction of 12.6 grams of magnesium with hydrochloric acid. 

Mg + 2HCl ===MgCl2 + HCl 

Calculate theoretical and percentage yield. 

Solution  

1 mole of Mg === 1 mole of MgCl2 

24 grams of Mg === 95 grams of MgCl2 

12.6 grams of Mg ==== X 

X = 12.6 x 95 / 24 = 49.88 grams of sodium chloride  

This is the theoretical yield 

Actual yield is 39 grams  

Percentage yield = Actual yield/Theoretical yield X 100 

% yield of the reaction = 39/49.88 X 100 

% yield of the reaction = 78.19% 

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