Chemical Equations and Reactions Solved Examples (Class 10 Chemistry)

Apply mole calculations to chemical equations and understand mass relationships in reactions. These foundational problems prepare for Class 11 stoichiometr

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TL;DR: Apply mole calculations to chemical equations and understand mass relationships in reactions. These foundational problems prepare for Class 11 stoichi…

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Apply mole calculations to chemical equations and understand mass relationships in reactions. These foundational problems prepare for Class 11 stoichiometr

Chemical Equations and Reactions — Solved Numerical Examples (Step by Step)

Example 1: Balance the equation: Fe + O₂ → Fe₂O₃ and find how many moles of Fe₂O₃ form from 4 moles of Fe.

Solution: Balancing: 4Fe + 3O₂ → 2Fe₂O₃
From equation: 4 mol Fe → 2 mol Fe₂O₃
4 mol Fe → 2 mol Fe₂O₃ (same ratio)
So 4 moles of Fe produce 2 moles of Fe₂O₃

Example 2: In the reaction 2H₂ + O₂ → 2H₂O, if 16 g of O₂ reacts, how many grams of H₂O are produced? (Atomic masses: H=1, O=16)

Solution: Molar mass O₂ = 32 g/mol, Molar mass H₂O = 18 g/mol
Moles of O₂ = 16 / 32 = 0.5 mol
From equation: 1 mol O₂ → 2 mol H₂O
0.5 mol O₂ → 1 mol H₂O
Mass of H₂O = 1 × 18 = 18 g

Example 3: Balance and solve: Mg + O₂ → MgO, then find mass of MgO from 24 g Mg. (Atomic masses: Mg=24, O=16)

Solution: Balancing: 2Mg + O₂ → 2MgO
Molar mass Mg = 24, Molar mass MgO = 24 + 16 = 40 g/mol
Moles of Mg = 24 / 24 = 1 mol
From equation: 2 mol Mg → 2 mol MgO
1 mol Mg → 1 mol MgO
Mass of MgO = 1 × 40 = 40 g

Example 4: In the reaction CH₄ + 2O₂ → CO₂ + 2H₂O, if 2 moles of CH₄ react, how many moles of O₂ are consumed?

Solution: From equation: 1 mol CH₄ requires 2 mol O₂
2 mol CH₄ require 2 × 2 = 4 mol O₂

Example 5: Balance: AgNO₃ + NaCl → AgCl + NaNO₃, then find moles of AgCl from 85 g AgNO₃. (Atomic masses: Ag=108, N=14, O=16, Na=23, Cl=35.5)

Solution: Balanced equation: AgNO₃ + NaCl → AgCl + NaNO₃
Molar mass AgNO₃ = 108 + 14 + 3(16) = 108 + 14 + 48 = 170 g/mol
Moles of AgNO₃ = 85 / 170 = 0.5 mol
From equation: 1 mol AgNO₃ → 1 mol AgCl
0.5 mol AgNO₃ → 0.5 mol AgCl

Example 6: If 44 g of CO₂ is completely formed, how many grams of C reacted? (Atomic masses: C=12, O=16; Reaction: C + O₂ → CO₂)

Solution: Molar mass CO₂ = 12 + 2(16) = 44 g/mol
Moles of CO₂ = 44 / 44 = 1 mol
From equation: 1 mol C → 1 mol CO₂
1 mol C required
Mass of C = 1 × 12 = 12 g

Tips

Always balance the equation using the smallest whole number coefficients.
Use molar ratios (coefficient ratios) from the balanced equation to relate moles of reactants and products.
Convert mass to moles, use stoichiometry, then convert back to mass if required.

Frequently Asked Questions

Why must chemical equations be balanced?

Balanced equations represent the law of conservation of mass—the number of atoms of each element is the same on both sides. This ensures stoichiometric calculations are correct.

Can you have fractional coefficients in a balanced equation?

In principle yes, but conventionally we use the smallest whole numbers. For example, 2H₂ + O₂ → 2H₂O is preferred over H₂ + 0.5O₂ → H₂O.

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