2
Question Question 2
When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Given,
When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
Now,
Carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8. Hence, 3 g of carbon will react with 50 g of oxygen but only 8 g of it will be used in producing 11 g of CO2 and 42 g [i.e., 50 g - 8 g] of oxygen will be left unused.
The above answer is governed by the law of constant proportions.
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Science | Chapter 3: Atoms and MoleculesWeb Content — Quick Reference
Chapter 3: Atoms and Molecules — Quick Reference
Quick Revision Points
- Law of conservation of mass: total mass unchanged in a reaction
- Law of constant proportions: elements in fixed mass ratio in a compound
- 1 u = 1/12 mass of C-12 = 1.66 × 10⁻²⁴ g
- Molecular mass = sum of atomic masses of all atoms in molecule
- 1 mole = 6.022 × 10²³ particles (Avogadro's number)
- Moles = mass / molar mass; Particles = moles × Nₐ
- Criss-cross method for writing chemical formulae using valencies
- Polyatomic ions: NH₄⁺, OH⁻, SO₄²⁻, NO₃⁻, CO₃²⁻, PO₄³⁻
Exam Tips for Chapter 3
- Read the detailed chapter notes for complete coverage of all NCERT topics.
- Practice all NCERT in-text and back exercise questions — they are frequently asked in exams.
- Focus on comparison tables, diagrams, and definitions — these are high-scoring areas.
- For numericals (if applicable), practice at least 20 problems of varying difficulty.
- Refer to the practice question bank (200+ questions) for thorough preparation.
Related Resources
- Detailed Notes: ch03-atoms-and-molecules.html
- Practice Questions: 100+ questions with answers in 05-practice-questions/
- Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
- Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html