CHEMICAL REACTIONS AND STOICHIOMETRY: Detailed Notes, Solved Examples & Exam Questions | Grade 10 Chemistry Unit 1

Hello my dear students! Welcome to your online chemistry class. Today, we are going to start a very important topic. Have you ever watched a piece of wood burning? Have you seen an iron nail getting rusty? What about the way a raw fruit becomes ripe and sweet? All these things happen because of chemical reactions. In this unit, we will learn exactly how these changes happen, how to write them down like a scientist, and how to balance them. This lesson is built directly from your Grade 10 Chemistry textbook to help you score high on your national and entrance exams. Are you ready? Let us dive in!

1.1 Introduction to Chemical Reactions

Change is everywhere around us. But in science, we do not just look at change; we sort it into two clear groups. Think about it. If you melt some ice to get water, is it still water? Yes! You just changed it from a solid to a liquid. You did not make a new substance. This is called a physical change. Another example is powdering a sugar cube. It is still sugar, just in smaller pieces.

But now, think about burning that same piece of wood. Do you get wood back? No. You get ash, smoke, and heat. The original substance is gone forever. New substances are formed. This is called a chemical change. A chemical change always happens because of a chemical reaction.

Let me ask you a quick question to check your understanding. How do you know a chemical reaction has happened? You look for clues! Your textbook gives us these clues: Formation of a new substance, change in color, change in temperature, production of heat or light. If you see these, a chemical reaction is taking place.

In a chemical reaction, the starting materials are called reactants. The new materials that are made at the end are called products. We write it like this: Reactants → Products. The arrow simply means “yield” or “produce”.

1.2 Chemical Equations

Scientists do not like writing long sentences to describe reactions. Imagine writing “One molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia gas” every single time! Instead, we use a shortcut called a chemical equation.

Steps to Write a Chemical Equation

Your textbook gives us three clear steps to follow. Let us look at them using the reaction between calcium carbonate and sulfuric acid.

1. Write a word equation: Calcium carbonate + Sulphuric acid → Calcium sulphate + Water + Carbon dioxide
2. Write the correct symbols and formulas: CaCO3 + H2SO4 → CaSO4 + H2O + CO2
3. Balance the equation: In this specific case, if you count the atoms on both sides, they are already equal! So it is balanced.

Physical State Symbols

This is a very common place to lose marks on your exam. After the formula, you must put a small letter in brackets to show what state the chemical is in. Here are the four you must memorize: (s) for Solid, (l) for Liquid, (g) for Gas, (aq) for Aqueous solution (dissolved in water).

So our equation from above becomes: \( \text{CaCO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{CaSO}_4\text{(s)} + \text{H}_2\text{O}\text{(l)} + \text{CO}_2\text{(g)} \)

1.2.2 Balancing Chemical Equations

Why do we balance equations? Think about the Law of Conservation of Mass. This law says that matter cannot be created or destroyed. If you start with 5 iron atoms, you must end with 5 iron atoms. You cannot just lose one! To show this on paper, we make sure the number of atoms on the left (reactants) is exactly equal to the number of atoms on the right (products). We do this by putting numbers in front of the formulas. These numbers are called coefficients.

Your textbook teaches you three different methods. Let us go through each one deeply so you can choose the one you like best.

Method A: The Inspection Method (Trial and Error)

This is the most common way. You just look at the equation and fix it step by step. Let us use the example from your book: Iron reacts with water to form iron(IV) oxide and hydrogen gas.

Step 1 & 2: \( \text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2 \) (Unbalanced)

Step 3: Make a table to count the atoms.

Step 4: Fix the unbalanced atoms. 1. Fix Fe: Put a 3 in front of Fe. Now we have 3 Fe on both sides. 2. Fix O: Put a 4 in front of H2O. Now we have 4 O on both sides. 3. Fix H: Now we have 4 H2O, which means 8 Hydrogen atoms. But we only have 2 H on the right! Put a 4 in front of H2 to make it 8.

Final Balanced Equation: \( 3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2 \)

Note: Never change the small subscripts in a formula to balance an equation! Only change the big coefficients in front.

Method B: The Least Common Multiple (LCM) Method

This method is very mathematical. It is great for harder equations with big polyatomic ions. Here is how it works: you look at the total valency (combining power) of the elements on each side, find their Least Common Multiple, and divide to get your coefficients.

Let us balance: Aluminum reacts with oxygen to form aluminum oxide (\( \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \)).

1. Write the valency above each element: 3    2×2=4    3×2=6    2×3=6
\( \text{Al} \)      \( \text{O}_2 \)        \( \text{Al}_2\text{O}_3 \)

2. Find the LCM of the total valencies (4 and 6). The LCM is 12.

3. Put 12 above the arrow. Divide 12 by the valency of each part.

Result: \( 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \) (Perfectly balanced!)

Method C: The Algebraic Method

This method uses letters like a, b, and c as placeholders for the coefficients. It is very powerful because it takes the guessing out of balancing. Let us use the book’s example: \( \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \).

1. Put letters in front: \( a\text{Al} + b\text{O}_2 \rightarrow c\text{Al}_2\text{O}_3 \)

2. Make math equations for each element: For Al: \( a = 2c \) For O: \( 2b = 3c \)

3. Pick the smallest letter to solve. Let us say \( a = 1 \). Then \( 1 = 2c \), so \( c = 1/2 \). Put \( c \) into the oxygen equation: \( 2b = 3(1/2) \), so \( 2b = 3/2 \), meaning \( b = 3/4 \).

4. We have fractions (1/2 and 3/4). We cannot have half a molecule in chemistry! So, find the lowest common multiple of the denominators (which is 4) and multiply all answers by 4. \( a = 1 \times 4 = 4 \) \( b = 3/4 \times 4 = 3 \) \( c = 1/2 \times 4 = 2 \)

Result: \( 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \). It matches perfectly!

1.3 Types of Chemical Reactions

Did you know that thousands of chemical reactions happen in your body right now while you are reading this? But do not worry! Scientists have grouped almost all chemical reactions into just four basic types. If you can learn these four types, you can predict the products of almost any reaction. Let us look at them using the summary box from your textbook.

A. Direct Combination Reactions (Synthesis)

This is when two or more things join together to make ONE new thing. Think of two people holding hands to form one circle. The key sign of a combination reaction is that there is always only one product.

Example from the book: \( 2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl} \). Two elements (Sodium and Chlorine) combine to make one compound (Salt). Your book also shows an experiment where Iron filings and Sulphur powder are heated together to form Iron Sulphide (FeS).

B. Decomposition Reactions (Analysis)

This is the exact opposite of combination. One compound breaks down into two or more simpler substances. To break a compound apart, you usually need to add energy. We show this by putting a triangle (the Greek letter Delta, \( \Delta \)) above the arrow to mean “heat is added”.

Examples from the book: \( 2\text{HgO} \xrightarrow{\Delta} 2\text{Hg} + \text{O}_2 \) (Mercury oxide breaks down into mercury metal and oxygen gas). \( 2\text{KClO}_3 \xrightarrow{\Delta} 2\text{KCl} + 3\text{O}_2 \) (Potassium chlorate breaks down).

C. Single Displacement Reactions

In this reaction, one lonely element kicks out another element from a compound. It is like a game of musical chairs. For this to work, the lone element must be more active (more reactive) than the element it is trying to kick out.

There are two main kinds you will see: 1. Metals displacing Hydrogen: Active metals like Zinc react with acids to push out hydrogen gas. \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \). (Note: Copper will NOT do this because it is not active enough!). 2. Metals displacing other Metals: Your book gives a great experiment. If you put an iron nail into Copper Sulphate solution (\( \text{CuSO}_4 \)), the iron is more active than copper. So, iron kicks copper out! The nail becomes coated in red-brown copper metal.

D. Double Displacement Reactions

This is like two couples dancing, and they swap partners! Two compounds trade their negative and positive ions with each other. A very common result of this reaction is the formation of a solid that falls out of the liquid. This solid is called a precipitate.

Book Example: \( \text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2\text{NaCl} \). The Barium and Sulphate ions found each other and formed Barium Sulphate. The downward arrow (\( \downarrow \)) means it is an insoluble precipitate.

1.4 Oxidation and Reduction Reactions (Redox)

Now we reach a topic that students sometimes find tricky, but I will make it very easy for you. Have you ever seen an old iron roof turning brown and flaky? That is rusting. Rusting is an example of an oxidation-reduction reaction, which we call Redox for short.

Let us define the terms simply. Think about the word “REDUX”. REDuction means gaining electrons (electrons are negative, so the charge reduces). OXidation means losing electrons.

These two always happen together. If one atom loses an electron, another atom must be right there to catch it!

1.4.1 Oxidation Number (Oxidation State)

Since we cannot actually see electrons moving, scientists invented a fake number called the Oxidation Number to help us track them. It is like a bookkeeping system. It is the charge an atom appears to have. There are 8 strict rules you must memorize from your textbook:

1. Rule 1: Any uncombined element by itself is 0. (e.g., Na = 0, O2 = 0, S8 = 0).
2. Rule 2: For a simple ion (one atom), it is exactly the charge. (e.g., Na+ is +1, Mg2+ is +2, S2- is -2).
3. Rule 3: Oxygen is usually -2. Exceptions: In peroxides (like Na2O2) it is -1. In superoxides (KO2) it is -1/2. In Oxygen difluoride (OF2) it is +2.
4. Rule 4: Hydrogen is usually +1. Exception: In metal hydrides (like NaH, CaH2) it is -1.
5. Rule 5: The sum of all oxidation numbers in a neutral compound is 0.
6. Rule 6: The sum of all oxidation numbers in a polyatomic ion equals the charge of the ion.
7. Rule 7: Group 1A metals are always +1. Group 2A metals are always +2.
8. Rule 8: The more electronegative element takes the negative number.

Let us practice a hard one from your book. Find the oxidation number of Mn in the permanganate ion, \( \text{MnO}_4^- \).

Let the oxidation number of Mn be \( x \). Oxygen is -2. There are 4 oxygens. The total charge is -1. \( x + 4(-2) = -1 \) \( x – 8 = -1 \) \( x = +7 \). So, Mn has an oxidation number of +7.

1.4.2 Oxidizing and Reducing Agents

Your textbook uses a very good example to explain this. Think about a wet dish and a towel. The dish is the “wetting agent” because it gives water. The towel is the “drying agent” because it takes water. In chemistry: The substance that LOSES electrons (gets oxidized) is the Reducing Agent. Why? Because by losing electrons, it causes something else to be reduced! The substance that GAINS electrons (gets reduced) is the Oxidizing Agent. Why? Because by grabbing electrons, it forces something else to be oxidized!

Exam Tip for Tests: Your book mentions some special color tests that examiners love to ask! – If an oxidizing agent like \( \text{KMnO}_4^- \) reacts, it turns from Purple to Colorless. – If \( \text{K}_2\text{Cr}_2\text{O}_7^{2-} \) reacts, it turns from Orange to Green. – If a reducing agent like KI reacts, it forms Iodine, which turns starch paper Blue-Black.

1.4.4 Balancing Redox Reactions: Oxidation-Number-Change Method

Sometimes, balancing by simply counting atoms is not enough for redox reactions. We must make sure the electrons lost equal the electrons gained! We use the oxidation numbers to do this. Let us balance the reaction from your book: \( \text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2 \).

Step 1: Assign oxidation numbers. \( \text{Fe}_2^{+3}\text{O}_3^{-2} + \text{C}^{+2}\text{O}^{-2} \rightarrow \text{Fe}^0 + \text{C}^{+4}\text{O}_2^{-2} \)

Step 2: Identify what changed. Fe goes from +3 to 0. (Decrease of 3 per atom. Since there are 2 Fe atoms, total decrease = 6). C goes from +2 to +4. (Increase of 2 per atom).

Step 3: Make the increase equal the decrease using coefficients. The LCM of 6 and 2 is 6. To make the Carbon increase equal 6, we multiply the Carbon change by 3. This means we put a 3 in front of CO and CO2! To make the Iron decrease equal 6, it is already 6, so we do not need to change the Fe2O3, but we must make sure there are 2 Fe on the right (which there already is).

Final Balanced Equation: \( \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \)

Exam Tips for Grade 10 Chemistry Students

My dear students, you have learned a lot today! Before you go, let me give you some final tips to help you score high on your national and entrance exams:

• Always check your states: Do not forget to write (s), (l), (g), or (aq) in your equations if the question asks for a complete equation.
• Memorize the diatomic gases: Hydrogen, Nitrogen, Oxygen, Fluorine, Chlorine, Bromine, Iodine. If they are alone, they must have a subscript of 2!
• Master the oxidation number rules: Especially the exceptions for Oxygen (-1 in peroxides) and Hydrogen (-1 in metal hydrides). Examiners love to trick you with Na2O2 or NaH!
• Remember OIL RIG: If you get confused about agents, just remember that the reducing agent is the one that gets Oxidized (Loses electrons).

Keep practicing the balancing methods. The more you do it, the faster you will become. Good luck with your studies!