Stoichiometry Solved Examples (Class 11 Chemistry)

Apply stoichiometric principles to balance equations and calculate mass/mole relationships in chemical reactions. These problems bridge mole concept to rea

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TL;DR: Apply stoichiometric principles to balance equations and calculate mass/mole relationships in chemical reactions. These problems bridge mole concept t…

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Apply stoichiometric principles to balance equations and calculate mass/mole relationships in chemical reactions. These problems bridge mole concept to rea

Stoichiometry — Solved Numerical Examples (Step by Step)

Example 1: In the reaction H₂ + Cl₂ → 2HCl, if 2 moles of H₂ react, how many moles of HCl are produced?

Solution: From the balanced equation: 1 mol H₂ → 2 mol HCl
2 mol H₂ → (2 × 2) = 4 mol HCl

Example 2: In the reaction 2Fe + 3Cl₂ → 2FeCl₃, if 56 g of Fe reacts, what mass of FeCl₃ is produced? (Atomic masses: Fe=56, Cl=35.5)

Solution: Molar mass of Fe = 56 g/mol, Molar mass of FeCl₃ = 56 + 3(35.5) = 56 + 106.5 = 162.5 g/mol
Moles of Fe = 56 / 56 = 1 mol
From equation: 2 mol Fe → 2 mol FeCl₃
1 mol Fe → 1 mol FeCl₃
Mass of FeCl₃ = 1 × 162.5 = 162.5 g

Example 3: In the reaction C + O₂ → CO₂, if 24 g of C reacts completely, what volume of CO₂ is produced at STP? (Atomic mass: C=12; Molar volume at STP = 22.4 L/mol)

Solution: Molar mass of C = 12 g/mol
Moles of C = 24 / 12 = 2 mol
From equation: 1 mol C → 1 mol CO₂
2 mol C → 2 mol CO₂
Volume of CO₂ = 2 × 22.4 = 44.8 L

Example 4: In the reaction CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, if 20 g of CaCO₃ reacts, how many moles of CO₂ are produced? (Atomic masses: Ca=40, C=12, O=16, H=1, Cl=35.5)

Solution: Molar mass of CaCO₃ = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g/mol
Moles of CaCO₃ = 20 / 100 = 0.2 mol
From equation: 1 mol CaCO₃ → 1 mol CO₂
0.2 mol CaCO₃ → 0.2 mol CO₂

Example 5: Balance the equation and find limiting reagent: 10 g Fe + 10 g Cl₂ → FeCl₃ (Atomic masses: Fe=56, Cl=35.5)

Solution: Balanced equation: 2Fe + 3Cl₂ → 2FeCl₃
Molar mass Fe = 56 g/mol, Molar mass Cl₂ = 71 g/mol
Moles of Fe = 10 / 56 = 0.179 mol
Moles of Cl₂ = 10 / 71 = 0.141 mol
From equation: 2 mol Fe needs 3 mol Cl₂
0.179 mol Fe needs (3/2) × 0.179 = 0.269 mol Cl₂
But we have only 0.141 mol Cl₂, so Cl₂ is the limiting reagent

Example 6: In the reaction Na + H₂O → NaOH + H₂, if 9.2 g of Na reacts, how many grams of NaOH are formed? (Atomic masses: Na=23, O=16, H=1)

Solution: Balanced equation: 2Na + 2H₂O → 2NaOH + H₂
Molar mass Na = 23 g/mol, Molar mass NaOH = 23 + 16 + 1 = 40 g/mol
Moles of Na = 9.2 / 23 = 0.4 mol
From equation: 2 mol Na → 2 mol NaOH
0.4 mol Na → 0.4 mol NaOH
Mass of NaOH = 0.4 × 40 = 16 g

Tips

Always balance the equation first before solving stoichiometry problems.
Use molar ratios from the balanced equation to convert between moles of reactants and products.
The limiting reagent is the one that runs out first and determines the maximum product formed.

Frequently Asked Questions

What is a limiting reagent?

In a reaction, the limiting reagent is the substance that is completely consumed and restricts the amount of product formed. The other reagents are in excess.

How do you identify the limiting reagent?

Calculate the number of moles of each reactant. Using stoichiometric ratios from the balanced equation, determine which reactant produces the least amount of product.

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