⚖️ Balanced Chemical Equation
संतुलित रासायनिक समीकरण — Class 10 CBSE
संतुलित रासायनिक समीकरण — Class 10 CBSE
1️⃣ Description / परिचय
English
A chemical equation represents a chemical reaction using symbols and formulae. It is balanced when the number of atoms of every element is the same on the reactant and product sides. According to the law of conservation of mass, matter is neither created nor destroyed during a chemical reaction.हिन्दी
रासायनिक समीकरण किसी रासायनिक अभिक्रिया को संकेतों तथा रासायनिक सूत्रों द्वारा प्रदर्शित करता है। यह समीकरण तब संतुलित कहलाता है जब प्रत्येक तत्व के परमाणुओं की संख्या अभिकारक तथा उत्पाद दोनों ओर समान हो। द्रव्यमान संरक्षण के नियम के अनुसार रासायनिक अभिक्रिया में पदार्थ न तो उत्पन्न होता है और न ही नष्ट होता है।
Golden Rule / मुख्य नियम:
Reactant atoms = Product atoms
अभिकारक पक्ष के परमाणु = उत्पाद पक्ष के परमाणु
✔ केवल coefficient बदलें।
❌ Subscript कभी न बदलें।
Reactant atoms = Product atoms
अभिकारक पक्ष के परमाणु = उत्पाद पक्ष के परमाणु
✔ केवल coefficient बदलें।
❌ Subscript कभी न बदलें।
2️⃣ Attractive Animation / Animation द्वारा समझें
Atoms are conserved / परमाणु संरक्षित रहते हैं
H
H
O
→
H₂O
H₂O
2H₂ + O₂ → 2H₂O
2H₂ + O₂ → 2H₂O
3️⃣ Step-by-Step Balancing Method / संतुलन की विधि
Step 1: Write the correct chemical formulae.
सही रासायनिक सूत्र लिखिए।
सही रासायनिक सूत्र लिखिए।
Step 2: Count atoms of every element.
प्रत्येक तत्व के परमाणु गिनिए।
प्रत्येक तत्व के परमाणु गिनिए।
Step 3: Balance the element appearing in the most complex compound.
जटिल यौगिक में उपस्थित तत्व को पहले संतुलित करें।
जटिल यौगिक में उपस्थित तत्व को पहले संतुलित करें।
Step 4: Balance hydrogen and oxygen near the end when convenient.
जहाँ उचित हो, H तथा O को अंत में संतुलित करें।
जहाँ उचित हो, H तथा O को अंत में संतुलित करें।
Step 5: Use the smallest whole-number coefficients.
सबसे छोटे पूर्णांक गुणांक लगाएँ।
सबसे छोटे पूर्णांक गुणांक लगाएँ।
Step 6: Count every atom again.
अंत में प्रत्येक परमाणु की पुनः जाँच करें।
अंत में प्रत्येक परमाणु की पुनः जाँच करें।
4️⃣ Fully Solved Example / पूर्ण हल
Question: Balance Fe + O₂ → Fe₂O₃
Step 1: Oxygen on left = 2, right = 3.
LCM of 2 and 3 = 6.
Put 3 before O₂ and 2 before Fe₂O₃:
हिन्दी: ऑक्सीजन को संतुलित करने के लिए 2 और 3 का LCM = 6 लिया। इसलिए 3O₂ तथा 2Fe₂O₃ लिया। अब Fe = 4 हो गया, इसलिए 4Fe लगाया।
LCM of 2 and 3 = 6.
Put 3 before O₂ and 2 before Fe₂O₃:
Fe + 3O₂ → 2Fe₂O₃
Now Fe on right = 4.
Therefore put 4 before Fe:
4Fe + 3O₂ → 2Fe₂O₃
Final Answer:
4Fe + 3O₂ → 2Fe₂O₃
हिन्दी: ऑक्सीजन को संतुलित करने के लिए 2 और 3 का LCM = 6 लिया। इसलिए 3O₂ तथा 2Fe₂O₃ लिया। अब Fe = 4 हो गया, इसलिए 4Fe लगाया।
5️⃣ 50 MCQs — CBSE Foundation + Competitive / NEET Level
6️⃣ CBSE 2 Marks Questions — 10 Questions
Q1. What is a balanced chemical equation? Balance H₂ + O₂ → H₂O.
2 Marks
Solution / हल:
Balanced equation वह है जिसमें दोनों ओर प्रत्येक तत्व के परमाणुओं की संख्या समान हो।
Oxygen = 2 on left and 1 on right.
Put 2 before H₂O:
Balanced equation वह है जिसमें दोनों ओर प्रत्येक तत्व के परमाणुओं की संख्या समान हो।
Oxygen = 2 on left and 1 on right.
Put 2 before H₂O:
H₂ + O₂ → 2H₂O
Hydrogen becomes 4, so put 2 before H₂.
2H₂ + O₂ → 2H₂O
Q2. Why can we not change subscripts while balancing an equation?
2 Marks
Changing a subscript changes the chemical formula and therefore changes
the identity of the substance.
Hindi: Subscript बदलने से पदार्थ का रासायनिक सूत्र और पहचान बदल जाती है। इसलिए केवल coefficient बदला जाता है।
Hindi: Subscript बदलने से पदार्थ का रासायनिक सूत्र और पहचान बदल जाती है। इसलिए केवल coefficient बदला जाता है।
Q3. Balance: Mg + O₂ → MgO.
2 Marks
O₂ contains 2 O atoms, so put 2 before MgO. Then put 2 before Mg.
2Mg + O₂ → 2MgO
Q4. State the law on which balancing of equations is based.
2 Marks
Law of Conservation of Mass.
द्रव्यमान संरक्षण का नियम।
Matter is neither created nor destroyed during a chemical reaction.
द्रव्यमान संरक्षण का नियम।
Matter is neither created nor destroyed during a chemical reaction.
Q5. Balance: Zn + HCl → ZnCl₂ + H₂.
2 Marks
There are two Cl atoms in ZnCl₂, so use 2HCl.
Zn + 2HCl → ZnCl₂ + H₂
Q6. What is the role of coefficient in a chemical equation?
2 Marks
A coefficient shows the relative number of molecules/formula units or
moles involved in a reaction.
यह अभिक्रिया में पदार्थों का stoichiometric ratio बताता है।
यह अभिक्रिया में पदार्थों का stoichiometric ratio बताता है।
Q7. Balance: N₂ + H₂ → NH₃.
2 Marks
N₂ requires 2NH₃. This gives 6 H atoms, so use 3H₂.
N₂ + 3H₂ → 2NH₃
Q8. Is CaCO₃ → CaO + CO₂ balanced? Give reason.
2 Marks
Yes.
Ca = 1, C = 1 and O = 3 on both sides. Therefore it is already balanced.
Ca = 1, C = 1 and O = 3 on both sides. Therefore it is already balanced.
Q9. What is meant by stoichiometric coefficient?
2 Marks
The number written before a chemical formula is called its
stoichiometric coefficient.
उदाहरण: 2H₂O में 2 coefficient है।
उदाहरण: 2H₂O में 2 coefficient है।
Q10. Balance: Al + O₂ → Al₂O₃.
2 Marks
O LCM = 6.
4Al + 3O₂ → 2Al₂O₃
Check: Al = 4 and O = 6 on both sides.
7️⃣ CBSE 3 Marks Questions — 10 Questions
Q1. Balance the equation: Fe + O₂ → Fe₂O₃ and explain the steps.
3 Marks
Step 1: O atoms = 2 and 3.
Step 2: LCM = 6.
Step 3: Put 3 before O₂ and 2 before Fe₂O₃.
Step 4: Fe becomes 4.
Step 2: LCM = 6.
Step 3: Put 3 before O₂ and 2 before Fe₂O₃.
Step 4: Fe becomes 4.
4Fe + 3O₂ → 2Fe₂O₃
Q2. Balance: CH₄ + O₂ → CO₂ + H₂O.
3 Marks
C is already 1.
H = 4, therefore put 2 before H₂O.
O on right = 4, therefore put 2 before O₂.
H = 4, therefore put 2 before H₂O.
O on right = 4, therefore put 2 before O₂.
CH₄ + 2O₂ → CO₂ + 2H₂O
Check: C = 1, H = 4, O = 4.
Q3. Explain why balancing is necessary in a chemical equation.
3 Marks
1. It follows conservation of mass.
2. Number of atoms remains unchanged.
3. It gives correct stoichiometric/mole ratios. Hindi: समीकरण का संतुलन द्रव्यमान संरक्षण, परमाणुओं के संरक्षण और सही mole ratio के लिए आवश्यक है।
2. Number of atoms remains unchanged.
3. It gives correct stoichiometric/mole ratios. Hindi: समीकरण का संतुलन द्रव्यमान संरक्षण, परमाणुओं के संरक्षण और सही mole ratio के लिए आवश्यक है।
Q4. Balance: Na + H₂O → NaOH + H₂.
3 Marks
Start with Na and O.
Use 2Na and 2H₂O.
Use 2Na and 2H₂O.
2Na + 2H₂O → 2NaOH + H₂
Check: Na = 2, O = 2, H = 4.
Q5. Differentiate between coefficient and subscript.
3 Marks
Coefficient: Written before formula; used for balancing.
Example: 2H₂O. Subscript: Written within formula; represents number of atoms.
Example: H₂O. Subscript बदलना substance बदल सकता है, coefficient बदलने से amount changes.
Example: 2H₂O. Subscript: Written within formula; represents number of atoms.
Example: H₂O. Subscript बदलना substance बदल सकता है, coefficient बदलने से amount changes.
Q6. Balance: KClO₃ → KCl + O₂.
3 Marks
O atoms = 3 and 2.
LCM = 6.
Use 2KClO₃ and 3O₂.
LCM = 6.
Use 2KClO₃ and 3O₂.
2KClO₃ → 2KCl + 3O₂
K and Cl automatically become balanced.
Q7. Balance: Ca + H₂O → Ca(OH)₂ + H₂.
3 Marks
Ca = 1.
Ca(OH)₂ contains 2 O, so use 2H₂O.
Ca(OH)₂ contains 2 O, so use 2H₂O.
Ca + 2H₂O → Ca(OH)₂ + H₂
Ca = 1, O = 2, H = 4 on both sides.
Q8. What does 2H₂ + O₂ → 2H₂O tell us?
3 Marks
It tells us that:
1. 2 molecules/moles H₂ react with 1 molecule/mole O₂.
2. 2 molecules/moles H₂O are formed.
3. H and O atoms are conserved.
1. 2 molecules/moles H₂ react with 1 molecule/mole O₂.
2. 2 molecules/moles H₂O are formed.
3. H and O atoms are conserved.
Q9. Balance: Fe₂O₃ + CO → Fe + CO₂.
3 Marks
Fe₂O₃ has 2 Fe, so use 2Fe.
Use 3CO to balance oxygen.
Use 3CO to balance oxygen.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
All Fe, C and O atoms are equal.
Q10. Give three rules for balancing chemical equations.
3 Marks
1. Never change chemical formulae.
2. Change only coefficients.
3. Verify every element after balancing.
2. Change only coefficients.
3. Verify every element after balancing.
8️⃣ CBSE 4 Marks Questions — 10 Questions
Q1. Balance C₂H₆ + O₂ → CO₂ + H₂O with complete verification.
4 Marks
C₂ requires 2CO₂.
H₆ requires 3H₂O.
O on right = 4 + 3 = 7 O atoms.
Therefore use 7/2 O₂. Multiply entire equation by 2.
C = 4, H = 12, O = 14 on both sides.
H₆ requires 3H₂O.
O on right = 4 + 3 = 7 O atoms.
Therefore use 7/2 O₂. Multiply entire equation by 2.
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Check:C = 4, H = 12, O = 14 on both sides.
Q2. Balance P + O₂ → P₂O₅ and explain the LCM method.
4 Marks
O = 2 on left and 5 on right.
LCM = 10.
Use 5O₂ and 2P₂O₅.
P on right = 4, so use 4P.
LCM = 10.
Use 5O₂ and 2P₂O₅.
P on right = 4, so use 4P.
4P + 5O₂ → 2P₂O₅
P = 4 and O = 10 on both sides.
Q3. Balance NH₃ + O₂ → NO + H₂O and verify.
4 Marks
N = 1, so choose 4NH₃ and 4NO.
H = 12, therefore 6H₂O.
O on right = 4 + 6 = 10.
Therefore 5O₂.
H = 12, therefore 6H₂O.
O on right = 4 + 6 = 10.
Therefore 5O₂.
4NH₃ + 5O₂ → 4NO + 6H₂O
N = 4, H = 12, O = 10 on both sides.
Q4. Explain the importance of coefficients in stoichiometry with an example.
4 Marks
Consider:
2H₂ + O₂ → 2H₂O
The coefficients give the mole ratio:
H₂ : O₂ : H₂O = 2 : 1 : 2.
Thus 2 mol H₂ reacts with 1 mol O₂ to form 2 mol H₂O.
Q5. Balance Al + HCl → AlCl₃ + H₂ by hit-and-trial method.
4 Marks
AlCl₃ contains 3 Cl. To make an even number of Cl, use 2AlCl₃.
Then use 2Al and 6HCl.
Hydrogen = 6, so use 3H₂.
2Al + 6HCl → 2AlCl₃ + 3H₂
All atoms are equal.
Q6. Explain four differences between balanced and unbalanced equations.
4 Marks
Balanced:
1. Atoms equal on both sides.
2. Obeys conservation of mass.
3. Gives correct mole ratio.
4. Suitable for quantitative calculations. Unbalanced:
Atoms are unequal and stoichiometric calculations are incorrect.
1. Atoms equal on both sides.
2. Obeys conservation of mass.
3. Gives correct mole ratio.
4. Suitable for quantitative calculations. Unbalanced:
Atoms are unequal and stoichiometric calculations are incorrect.
Q7. Balance BaCl₂ + Na₂SO₄ → BaSO₄ + NaCl and explain.
4 Marks
Ba = 1 and SO₄ = 1 already.
Cl₂ requires 2NaCl.
Na₂ is then also balanced.
Cl₂ requires 2NaCl.
Na₂ is then also balanced.
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl
Ba = 1, Cl = 2, Na = 2, SO₄ = 1.
Q8. Balance H₂S + O₂ → SO₂ + H₂O with verification.
4 Marks
Use 2H₂S and 2SO₂ to balance S.
H becomes 4, so use 2H₂O.
O on right = 6, so use 3O₂.
H becomes 4, so use 2H₂O.
O on right = 6, so use 3O₂.
2H₂S + 3O₂ → 2SO₂ + 2H₂O
H = 4, S = 2, O = 6 on both sides.
Q9. Explain why changing H₂O to H₂O₂ is incorrect during balancing.
4 Marks
H₂O and H₂O₂ are different substances.
Changing H₂O to H₂O₂ changes the number of oxygen atoms in the
molecule and therefore changes the chemical identity.
Correct method:
2H₂ + O₂ → 2H₂O
Not by changing H₂O into H₂O₂.
Q10. Balance Fe + H₂O → Fe₃O₄ + H₂.
4 Marks
Fe₃O₄ contains 3Fe and 4O.
Use 3Fe and 4H₂O.
Hydrogen becomes 8, so use 4H₂.
Use 3Fe and 4H₂O.
Hydrogen becomes 8, so use 4H₂.
3Fe + 4H₂O → Fe₃O₄ + 4H₂
9️⃣ CBSE 5 Marks Questions — 10 Questions
Q1. Explain the complete procedure for balancing a chemical equation with example Fe + O₂ → Fe₂O₃.
5 Marks
Step 1: Write the equation.
Fe + O₂ → Fe₂O₃ Step 2: Count atoms.
O = 2 left, 3 right. Step 3: Take LCM = 6. Step 4: Use 3O₂ and 2Fe₂O₃. Step 5: Balance Fe by using 4Fe.
Fe = 4 on both sides.
O = 6 on both sides. Hence balanced.
Fe + O₂ → Fe₂O₃ Step 2: Count atoms.
O = 2 left, 3 right. Step 3: Take LCM = 6. Step 4: Use 3O₂ and 2Fe₂O₃. Step 5: Balance Fe by using 4Fe.
4Fe + 3O₂ → 2Fe₂O₃
Verification:Fe = 4 on both sides.
O = 6 on both sides. Hence balanced.
Q2. Balance C₃H₈ + O₂ → CO₂ + H₂O and explain every step.
5 Marks
C = 3, so use 3CO₂.
H = 8, so use 4H₂O.
O on product side = 6 + 4 = 10.
Therefore use 5O₂.
C = 3, H = 8, O = 10 on both sides.
H = 8, so use 4H₂O.
O on product side = 6 + 4 = 10.
Therefore use 5O₂.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Verification:C = 3, H = 8, O = 10 on both sides.
Q3. Explain the law of conservation of mass and connect it with balancing.
5 Marks
The law states that mass can neither be created nor destroyed in a
chemical reaction.
Since atoms are conserved, the number of atoms of each element must
remain unchanged.
Therefore a chemical equation must be balanced.
Example:
2H₂ + O₂ → 2H₂O
H = 4 and O = 2 on both sides.
Thus the equation obeys conservation of mass.
Q4. Balance: KMnO₄ + HCl → KCl + MnCl₂ + H₂O + Cl₂.
5 Marks
This is a higher-level redox balancing problem.
Use:
K = 2
Mn = 2
O = 8
H = 16
Cl = 16 All are equal on both sides.
2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂
Verification:K = 2
Mn = 2
O = 8
H = 16
Cl = 16 All are equal on both sides.
Q5. Balance: Fe₂O₃ + CO → Fe + CO₂ and explain its importance in metallurgy.
5 Marks
Fe₂O₃ contains 2 Fe, therefore use 2Fe.
There are 3 O atoms, so use 3CO.
This gives 3CO₂.
There are 3 O atoms, so use 3CO.
This gives 3CO₂.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
This reaction represents reduction of iron(III) oxide by carbon monoxide
during extraction of iron.
Q6. Explain five common mistakes students make while balancing equations.
5 Marks
1. Changing subscripts.
2. Forgetting to count atoms again.
3. Using fractional coefficients without simplifying.
4. Balancing only one element.
5. Ignoring polyatomic groups when they remain unchanged. Exam Tip: Always perform a final atom count.
2. Forgetting to count atoms again.
3. Using fractional coefficients without simplifying.
4. Balancing only one element.
5. Ignoring polyatomic groups when they remain unchanged. Exam Tip: Always perform a final atom count.
Q7. Balance C₂H₆ + O₂ → CO₂ + H₂O using fractional coefficient first and then convert to whole numbers.
5 Marks
C₂ → 2CO₂.
H₆ → 3H₂O.
O required = 7 atoms = 7/2 O₂. So:
H₆ → 3H₂O.
O required = 7 atoms = 7/2 O₂. So:
C₂H₆ + 7/2O₂ → 2CO₂ + 3H₂O
Multiply the complete equation by 2:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
This is the smallest whole-number form.
Q8. Explain the relation between balanced equations and mole ratios.
5 Marks
Consider:
2H₂ + O₂ → 2H₂O
Coefficient ratio:
H₂ : O₂ : H₂O = 2 : 1 : 2.
Therefore:
2 mol H₂ react with 1 mol O₂ and produce 2 mol H₂O.
This is why balancing is essential before stoichiometric calculations.
Q9. Balance NH₃ + O₂ → NO + H₂O and verify conservation of atoms.
5 Marks
Balance N first:
4NH₃ → 4NO.
H = 12, therefore:
6H₂O.
O on product side = 10, therefore:
5O₂.
Final:
4NH₃ + 5O₂ → 4NO + 6H₂O
N = 4, H = 12, O = 10 on both sides.
Q10. Describe how you would check whether any given chemical equation is balanced.
5 Marks
Step 1: List all elements.
Step 2: Count each atom on reactant side.
Step 3: Count each atom on product side.
Step 4: Compare the numbers.
Step 5: If all are equal, equation is balanced. Example:
O = 2 on both sides. Therefore balanced.
Step 2: Count each atom on reactant side.
Step 3: Count each atom on product side.
Step 4: Compare the numbers.
Step 5: If all are equal, equation is balanced. Example:
2Mg + O₂ → 2MgO
Mg = 2 on both sides.O = 2 on both sides. Therefore balanced.
🔟 CBSE 6 Marks Questions — 10 Questions
Q1. Explain the complete concept of balancing chemical equations and solve three examples.
6 Marks
Concept: A balanced equation has equal atoms of every element on
both sides.
Example 1:
2H₂ + O₂ → 2H₂O
Example 2:
4Fe + 3O₂ → 2Fe₂O₃
Example 3:
CH₄ + 2O₂ → CO₂ + 2H₂O
In all cases the number of atoms of each element remains conserved.
Hindi: संतुलन का मुख्य उद्देश्य प्रत्येक तत्व के परमाणुओं को दोनों
पक्षों पर समान करना है।
Q2. Balance C₂H₆ + O₂ → CO₂ + H₂O and calculate the mole ratio of all substances.
6 Marks
Balance:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Therefore mole ratio:
C₂H₆ : O₂ : CO₂ : H₂O = 2 : 7 : 4 : 6
Thus 2 mol ethane reacts with 7 mol oxygen to produce 4 mol CO₂
and 6 mol water.
Verification:
C = 4, H = 12, O = 14 on both sides.
Q3. Explain the law of conservation of mass with a balanced chemical equation and atom table.
6 Marks
Consider:
Therefore no atom is created or destroyed.
Hence the equation obeys the law of conservation of mass.
Hindi: प्रत्येक तत्व के परमाणुओं की संख्या समान होने के कारण
द्रव्यमान संरक्षित रहता है।
2H₂ + O₂ → 2H₂O
| Element | Reactant Atoms | Product Atoms |
|---|---|---|
| H | 4 | 4 |
| O | 2 | 2 |
Q4. Balance Fe₂O₃ + CO → Fe + CO₂ and explain its role in extraction of iron.
6 Marks
Fe₂O₃ contains 2 Fe atoms.
Therefore:
Fe₂O₃ → 2Fe
O = 3, therefore use 3CO.
Carbon then requires 3CO₂.
Final equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
CO acts as the reducing agent and Fe₂O₃ is reduced to Fe.
Hindi: इस अभिक्रिया में CO, Fe₂O₃ से oxygen हटाता है और iron प्राप्त होता है।
Q5. Balance KMnO₄ + HCl → KCl + MnCl₂ + H₂O + Cl₂ and verify every element.
6 Marks
Balanced equation:
Hence the equation is completely balanced.
2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂
Verification:
| Element | Left | Right |
|---|---|---|
| K | 2 | 2 |
| Mn | 2 | 2 |
| O | 8 | 8 |
| H | 16 | 16 |
| Cl | 16 | 16 |
Q6. Explain the hit-and-trial method of balancing with suitable examples.
6 Marks
The hit-and-trial method involves changing coefficients systematically
until every element becomes equal.
Example:
Fe + O₂ → Fe₂O₃
First balance oxygen using LCM 6:
Fe + 3O₂ → 2Fe₂O₃
Now Fe = 4 on right:
4Fe + 3O₂ → 2Fe₂O₃
Verification:
Fe = 4 and O = 6 on both sides.
Therefore balanced.
Q7. Explain five rules of balancing chemical equations with examples.
6 Marks
Rule 1: Write correct formulae.
Rule 2: Count every atom.
Rule 3: Change coefficients only.
Rule 4: Use smallest whole-number ratio.
Rule 5: Recheck every element. Example:
Rule 2: Count every atom.
Rule 3: Change coefficients only.
Rule 4: Use smallest whole-number ratio.
Rule 5: Recheck every element. Example:
2Na + 2H₂O → 2NaOH + H₂
Never change H₂O into another chemical formula.
Hindi: सही सूत्र, परमाणु गणना, केवल coefficient परिवर्तन,
छोटे पूर्णांक तथा final verification — ये पाँच मुख्य नियम हैं।
Q8. Balance C₃H₈ + O₂ → CO₂ + H₂O and use the equation to explain stoichiometry.
6 Marks
Balanced equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Mole ratio:
1 : 5 : 3 : 4
Therefore:
1 mol C₃H₈ requires 5 mol O₂ and produces 3 mol CO₂
and 4 mol H₂O.
This demonstrates how balanced equations provide quantitative
relationships between substances.
Hindi: संतुलित समीकरण से पदार्थों के बीच mole ratio प्राप्त होता है।
Q9. A student balances H₂ + O₂ → H₂O as H₂ + O₂ → H₂O. Identify the error and correct it.
6 Marks
Error:
Hydrogen atoms:
Left = 2
Right = 2 Oxygen atoms: Left = 2
Right = 1 Therefore oxygen is not balanced. Put 2 before H₂O:
Right = 2 Oxygen atoms: Left = 2
Right = 1 Therefore oxygen is not balanced. Put 2 before H₂O:
H₂ + O₂ → 2H₂O
Now H becomes 4 on right.
Therefore put 2 before H₂:
2H₂ + O₂ → 2H₂O
Final equation is balanced.
Q10. Write a complete CBSE-style answer explaining why a chemical equation must always be balanced.
6 Marks
Model CBSE Answer:
A chemical equation must always be balanced because it represents a
chemical reaction in which atoms are neither created nor destroyed.
According to the law of conservation of mass, the total number of atoms
of each element must remain unchanged.
Therefore the number of atoms of every element on the reactant side
must equal the number on the product side.
For example:
2H₂ + O₂ → 2H₂O
Here H = 4 and O = 2 on both sides.
Hence the equation is balanced and obeys the law of conservation of
mass.
Hindi:
रासायनिक समीकरण को संतुलित करना आवश्यक है क्योंकि अभिक्रिया के दौरान
परमाणु न तो उत्पन्न होते हैं और न ही नष्ट होते हैं। द्रव्यमान संरक्षण
के नियम के कारण प्रत्येक तत्व के परमाणुओं की संख्या दोनों पक्षों पर
समान होनी चाहिए।
1️⃣1️⃣ One-Page Quick Revision / परीक्षा से पहले त्वरित Revision
Balanced Equation = Equal atoms of every element on both sides
✔ Law of Conservation of Mass
✔ Count atoms carefully
✔ Change only coefficients
✔ Never change subscripts
✔ Use smallest whole-number coefficients
✔ Check every element
✔ Balanced equation gives mole ratio
✔ Use balanced equation before numerical calculations
✔ Count atoms carefully
✔ Change only coefficients
✔ Never change subscripts
✔ Use smallest whole-number coefficients
✔ Check every element
✔ Balanced equation gives mole ratio
✔ Use balanced equation before numerical calculations
CBSE Answer Writing Tip:
For 3, 4, 5 and 6 marks questions, do not write only the final equation.
Show the balancing steps, atom counting and final verification.
CBSE परीक्षा Tip: केवल final answer लिखने के बजाय balancing के steps, atom counting और final verification जरूर लिखें। इससे step-wise marks प्राप्त करने में मदद मिलती है।
CBSE परीक्षा Tip: केवल final answer लिखने के बजाय balancing के steps, atom counting और final verification जरूर लिखें। इससे step-wise marks प्राप्त करने में मदद मिलती है।
1️⃣2️⃣ Final Exam Formula / अंतिम सूत्र
⚖️ REACTANT ATOMS = PRODUCT ATOMS
अभिकारक पक्ष के परमाणु = उत्पाद पक्ष के परमाणु
Only Coefficients Change ✅
Subscripts Never Change ❌
अभिकारक पक्ष के परमाणु = उत्पाद पक्ष के परमाणु
Only Coefficients Change ✅
Subscripts Never Change ❌
