balance eq 2

 Certainly, Mayank! Here's a well-structured, student-friendly, and CBSE Class 10–oriented explanation of Section 1.1.2: Balanced Chemical Equations, rewritten in a way that makes it easier to understand, exam-oriented, and ready for classroom delivery or student handouts.

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⚖️ 1.1.2 – Balanced Chemical Equations

(From Chapter: Chemical Reactions and Equations – Class 10)

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🎯 Why Do We Balance Equations?

According to the Law of Conservation of Mass:

“Mass can neither be created nor destroyed in a chemical reaction.”

✔️ That means:
👉 The number of atoms of each element must be the same on both sides of a chemical equation.

So, we must balance equations to obey this law.

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🔄 What is a Balanced Chemical Equation?

A balanced equation has the same number of each type of atom on both sides (reactants and products) of the arrow.

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📘 Example – From Activity 1.3

Word Equation:
Zinc + Sulphuric acid → Zinc sulphate + Hydrogen

Chemical Equation:

$$\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2$$

Let’s count atoms:

Element	LHS (Reactants)	RHS (Products)
Zn	1	1
H	2	2
S	1	1
O	4	4

✅ All atoms are balanced. So this is a balanced chemical equation.

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🔍 Balancing an Unbalanced Equation – Step-by-Step

Let’s balance this unbalanced equation:

$$\text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$$

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🔢 Step I: Draw boxes around each formula (no changes inside the box).

$$\boxed{\text{Fe}} + \boxed{\text{H}_2\text{O}} \rightarrow \boxed{\text{Fe}_3\text{O}_4} + \boxed{\text{H}_2}$$

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🔢 Step II: Count the atoms on both sides.

Element	LHS	RHS
Fe	1	3 ❌
H	2	2 ✅
O	1	4 ❌

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🔢 Step III: Start with the compound that has the most atoms (Fe₃O₄), balance oxygen first.

There are 4 oxygen atoms on the RHS, but only 1 on LHS.

👉 Multiply H₂O by 4:

$$\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$$

Now count again:

• O: 4 on both sides ✅

• H: 8 (in 4H₂O) on LHS, 2 on RHS ❌

• Fe: Still 1 vs 3 ❌

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🔢 Step IV: Balance Hydrogen

On LHS: 8 H atoms
On RHS: Only 2

👉 Multiply H₂ by 4:

$$\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$$

✅ Now H: 8 on both sides

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🔢 Step V: Balance Iron (Fe)

On RHS: 3 Fe atoms (in Fe₃O₄)
👉 Multiply Fe on LHS by 3:

$$3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$$

✅ All atoms are now balanced!

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✅ Balanced Equation:

$$\boxed{3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2}$$

This method is called the hit-and-trial method because we try different coefficients to balance the equation.

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🧪 Adding Physical States to the Balanced Equation

To make equations more informative, we show the physical state of each substance:

Symbol	Meaning
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous solution

Final Equation with States:

$$3\text{Fe (s)} + 4\text{H}_2\text{O (g)} \rightarrow \text{Fe}_3\text{O}_4 (s) + 4\text{H}_2 (g)$$

📝 (Water is in steam form, so we write H₂O as a gas.)

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🔺 Additional Notes:

Sometimes we also show conditions of the reaction (like temperature, pressure, catalyst) above or below the arrow.

Example:

$$\text{CO(g)} + 2\text{H}_2(g) \xrightarrow{340\,atm} \text{CH}_3\text{OH(l)}$$

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🧠 Now Try This (Homework or Class Activity)

Can you balance this?

$$\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3$$

✅ (Answer: 4Al + 3O₂ → 2Al₂O₃)

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🎤 Closing Line for Students:

“Balancing an equation is like making nature fair — every atom must be counted, and nothing is lost. That’s chemistry's way of keeping balance.”

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Would you like a step-by-step worksheet, practice quiz, or animated balancing tool to help students master this process visually?