Stoichiometry (SAQs)

Chemistry-1 | 5. Stoichiometry – SAQs:
Welcome to SAQs in Chapter 5: Stoichiometry. This page features the key FAQs for Short Answer Questions. Answers are given in simple English, with a Telugu explanation, and formatted according to the exam style. This will support your preparation and help you aim for top marks in your final exams.

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SAQ-1 : Balance the following redox equation in acidic medium by ion-electron method: Fe(aq)²⁺+Cr₂O₇aq₂→Fe_(aq)³⁺ + Cr(aq)³⁺

Introduction

Balancing redox reactions, especially in acid or base solutions, is a fundamental aspect of chemistry. The ion-electron method is an effective step-by-step approach for this purpose. It involves separating the oxidation and reduction half-reactions, balancing them independently, and then combining them. We will apply this method to balance the given redox equation in an acidic medium.

Balancing the Redox Reaction

1. Identify the Oxidation and Reduction Half Reactions:
   • Oxidation half-reaction: $$Fe^{+2} \rightarrow Fe^{+3}$$
   • Reduction half-reaction: $$Cr_2O_7^{−2} \rightarrow Cr^{+3}$$
2. Balance Atoms Other than Oxygen and Hydrogen:
   • Oxidation half-reaction remains the same.
   • Reduction half-reaction: $$Cr_2O_7^{−2} \rightarrow 2Cr^{+3}$$
3. Balance Oxygen Atoms Using Water:
   • Oxidation half-reaction: No change (no oxygen atoms present).
   • Reduction half-reaction: $$Cr_2O_7^{−2} \rightarrow 2Cr^{+3} + 7H_2O$$
4. Balance Hydrogen Atoms Using H⁺ Ions:
   • Oxidation half-reaction: No change (no hydrogen atoms present).
   • Reduction half-reaction: $$Cr_2O_7^{−2} + 14H^+ \rightarrow 2Cr^{+3} + 7H_2O$$
5. Balance Charge Using Electrons:
   • Oxidation half-reaction: $$Fe^{+2} \rightarrow Fe^{+3} + e^-$$
   • Reduction half-reaction: $$Cr_2O_7^{−2} + 14H^+ + 6e^- \rightarrow 2Cr^{+3} + 7H_2O$$
6. Equalize Electrons and Add the Half-Reactions Together:
   • Oxidation half-reaction (multiplied by 6): $$6Fe^{+2} \rightarrow 6Fe^{+3} + 6e^-$$
   • Combined equation: $$6Fe^{+2} + Cr_2O_7^{−2} + 14H^+ \rightarrow 6Fe^{+3} + 2Cr^{+3} + 7H_2O$$

Summary

The balanced redox equation in an acidic medium is $$6Fe^{+2} + Cr_2O_7^{−2} + 14H^+ \rightarrow 6Fe^{+3} + 2Cr^{+3} + 7H_2O$$ This equation exemplifies the ion-electron method’s systematic approach to balancing redox reactions, ensuring mass and charge conservation.

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SAQ-2 : Balance the following redox reaction by ion-electron method in acid medium: MnO₄(aq) +SO₂(g)→Mn(aq)⁺² + HSO₄(aq)^–

Introduction

Balancing redox reactions, particularly in acidic solutions, can be complex. The ion-electron method offers a systematic approach to achieve this. We will apply this method to balance the given redox reaction involving $$MnO_4^-$$​ and SO₂​ in an acidic medium.

Skeleton Equation: $$MnO_4^- + SO_2 \rightarrow Mn^{+2} + HSO_4^-$$

1. Identify the Oxidation and Reduction Half Reactions:
   • Oxidation half-reaction: $$SO_2 \rightarrow HSO_4^-$$
   • Reduction half-reaction: $$MnO_4^- \rightarrow Mn^{+2}$$
2. Balance Atoms Other than Oxygen and Hydrogen:
   • Oxidation half-reaction: $$SO_2 \rightarrow HSO_4^-$$
   • Reduction half-reaction: $$MnO_4^- \rightarrow Mn^{+2}$$
3. Balance Oxygen Atoms Using Water:
   • Oxidation half-reaction: $$SO_2 + 2H_2O \rightarrow HSO_4^-$$
   • Reduction half-reaction: $$MnO_4^- \rightarrow Mn^{+2} + 4H_2O$$
4. Balance Hydrogen Atoms Using H⁺ Ions:
   • Oxidation half-reaction: $$SO_2 + 2H_2O \rightarrow HSO_4^- + 4H^+$$
   • Reduction half-reaction: $$MnO_4^- + 8H^+ \rightarrow Mn^{+2} + 4H_2O$$
5. Balance Charge Using Electrons:
   • Oxidation half-reaction: $$SO_2 + 2H_2O \rightarrow HSO_4^- + 4H^+ + 2e^-$$
   • Reduction half-reaction: $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{+2} + 4H_2O$$

Equalize Electrons and Add the Half-Reactions Together:

1. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
   $$5(SO_2 + 2H_2O) \rightarrow 5HSO_4^- + 20H^+ + 10e^-$$
   $$2(MnO_4^- + 8H^+ + 5e^-) \rightarrow 2Mn^{+2} + 8H_2O$$
2. Combined equation: $$2MnO_4^- + 5SO_2 + 6H_2O \rightarrow 2Mn^{+2} + 5HSO_4^- + 16H^+$$
3. Final Balanced Redox Reaction in Acidic Medium: $$2MnO_4^- + 5SO_2 + 6H_2O \rightarrow 2Mn^{+2} + 5HSO_4^- + 16H^+$$

Summary:

This balanced equation demonstrates the ion-electron method for balancing redox reactions in acidic media. It’s important to note that water

(H₂O) and protons (H⁺) may sometimes cancel each other out, potentially simplifying the final equation. Checking for these at the end of the balancing process is a crucial step.

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SAQ-3 : Balance the following redox reaction by ion-electron method in acid medium: Cr₂O – 7(aq)^-2 + SO³(aq^-2→Cr(aq)⁺³ +SO₄(aq)^-2

Introduction

Balancing redox reactions in an acidic medium can be efficiently achieved using the ion-electron method. This method involves balancing the oxidation and reduction half-reactions separately before combining them. We’ll apply this approach to the reaction between $$Cr_2O_7^{2-}$$​ and $$SO_3^{2-}$$​ to form $$Cr^{3+}$$ and $$SO_4^{2-}$$

Balancing the Redox Reaction

1. Skeleton Equation: $$Cr_2O_7^{2-} + SO_3^{2-} \rightarrow Cr^{3+} + SO_4^{2-}​$$
2. Identify Oxidation and Reduction Half Reactions:
   • Oxidation half-reaction: $$SO_3^{2-} \rightarrow SO_4^{2-}​$$
   • Reduction half-reaction: $$Cr_2O_7^{2-} \rightarrow Cr^{3+}$$
3. Balance Atoms Other than Oxygen and Hydrogen:
   • Oxidation half-reaction remains unchanged.
   • Reduction half-reaction remains unchanged.
4. Balance Oxygen Atoms Using Water:
   • Oxidation half-reaction: $$SO_3^{2-} \rightarrow SO_4^{2-}$$
   • Reduction half-reaction: $$Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O$$
5. Balance Hydrogen Atoms Using H⁺ Ions:
   • Oxidation half-reaction: $$SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+$$
   • Reduction half-reaction remains the same.
6. Balance Charge Using Electrons:
   • Oxidation half-reaction: $$SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-$$
   • Reduction half-reaction: $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$
7. Equalize Electrons and Add the Half-Reactions Together:
   • Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1.
   • Combined equation: $$3(SO_3^{2-} + H_2O) + Cr_2O_7^{2-} + 14H^+ \rightarrow 3SO_4^{2-} + 6H^+ + 2Cr^{3+} + 7H_2O$$

Final Balanced Redox Reaction in Acidic Medium: $$Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$$

Summary

The balanced redox reaction in an acidic medium is This equation showcases the application of the ion-electron method to balance redox reactions. Each step, from identifying half-reactions to balancing charge and atoms, is critical for achieving a balanced equation that conserves mass and charge. Understanding this process is essential for students studying redox chemistry, as it provides a clear method for balancing complex redox equations.

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SAQ-4 : Balance the following equation in Basic medium by ion-electron method: MnO⁴ + I→MnO₂ + I₂

Introduction

Balancing redox reactions in a basic medium requires a careful approach. The ion-electron method is a systematic technique used for this purpose. We will apply this method to balance the reaction between $$MnO_4^-$$ and I⁻ to form MnO₂​ and I₂​.

Balancing the Redox Reaction

1. Skeleton Equation: $$MnO_4^- + I^- \rightarrow MnO_2 + I_2$$
2. Identify Oxidation and Reduction Half Reactions:
   • Oxidation half-reaction: $$I^- \rightarrow I_2$$
   • Reduction half-reaction: $$MnO_4^- \rightarrow MnO_2$$
3. Balance Atoms Other than Oxygen and Hydrogen:
   • Oxidation half-reaction: $$2I^- \rightarrow I_2$$
   • Reduction half-reaction remains unchanged.
4. Balance Oxygen Atoms Using Water Molecules:
   • Reduction half-reaction: $$MnO_4^- \rightarrow MnO_2 + 2H_2O$$
5. Balance Hydrogen Atoms Using Hydroxide Ions:
   • Reduction half-reaction: $$MnO_4^- + 4OH^- \rightarrow MnO_2 + 2H_2O + 4e^-$$
6. Balance Charge Using Electrons:
   • Oxidation half-reaction: $$2I^- \rightarrow I_2 + 2e^-$$
   • Reduction half-reaction already balanced for charge.
7. Equalize Electrons and Add the Half-Reactions Together:
   • Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1.
   • Combined equation: $$2MnO_4^- + 8OH^- + 4I^- \rightarrow 2MnO_2 + 4H_2O + 2I_2 + 8e^-$$

Final Balanced Redox Reaction in Basic Medium: $$2MnO_4^- + 4I^- + 8OH^- \rightarrow 2MnO_2 + 2I_2 + 4H_2O$$

Summary

The balanced redox reaction in a basic medium is $$2MnO_4^- + 4I^- + 8OH^- \rightarrow 2MnO_2 + 2I_2 + 4H_2O$$ This reaction is an excellent example of the ion-electron method applied in a basic environment, illustrating the importance of balancing both the mass and charge through each step of the process. Understanding this balancing technique is vital for students, as it equips them with the skills to address a range of redox reactions in different chemical environments.

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SAQ-5 : A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Introduction

Determining the empirical and molecular formulas of a compound is a fundamental aspect of chemistry. This involves analyzing the compound’s composition and its molar mass. We will calculate these formulas for a compound composed of hydrogen, carbon, and chlorine.

Determining the Empirical Formula

1. Given Compound Composition:
   • Hydrogen (H): 4.07%
   • Carbon (C): 24.27%
   • Chlorine (Cl): 71.65%
2. Convert Percentages to Moles:
   • Moles of H: $$\frac{4.07\text{ g}}{1.01\text{ g/mol}} \approx 4.03 \text{ moles}$$
   • Moles of C: $$\frac{24.27\text{ g}}{12.01\text{ g/mol}} \approx 2.02 \text{ moles}$$
   • Moles of Cl: $$\frac{71.65\text{ g}}{35.45\text{ g/mol}} \approx 2.02 \text{ moles}$$
3. Determine the Simplest Whole Number Ratio: Divide all mole values by the smallest number of moles: H:C:Cl ratio becomes approximately 2:1:1.
4. Empirical Formula: Based on the ratio, the empirical formula is H₂​CCl.

Determining the Molecular Formula

1. Molar Mass of the Compound: Given molar mass: 98.96 g/mol.
2. Molar Mass of the Empirical Formula: Calculated as: 2 (from H) + 12 (from C) + 35.5 (from Cl) = 49.5 g/mol.
3. Divide Molar Masses to Find Ratio: Divide the given molar mass by the empirical formula’s molar mass: $$\frac{98.96\text{ g/mol}}{49.5\text{ g/mol}} \approx 2$$
4. Molecular Formula: Multiply the empirical formula by the ratio (2): (H₂​CCl)₂​ simplifies to C₂​H₄​Cl₂​.

Summary

The empirical formula of the compound is H₂​CCl and the molecular formula is C₂​H₄​C_l2​. This process demonstrates the essential steps in deriving empirical and molecular formulas from given composition percentages and molar mass.

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SAQ-6 : A carbon compound contains 12.8%carbon, 2.1% hydrogen, 85.1% bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula.

Introduction

Calculating the molecular formula of a compound requires understanding its composition and molar mass. We will determine the molecular formula for a compound containing carbon, hydrogen, and bromine, given its percentage composition and molecular weight.

Empirical Formula Determination

1. Given Compound Composition:
   • Carbon (C): 12.8%
   • Hydrogen (H): 2.1%
   • Bromine (Br): 85.1%
2. Convert Percentages to Moles:
   • Moles of C: $$\frac{12.8\%}{12.01\, \text{g/mol}}$$
   • Moles of H: $$\frac{2.1\%}{1.008\, \text{g/mol}}$$
   • Moles of Br: $$\frac{85.1\%}{79.904\, \text{g/mol}}$$
3. Determine Simplest Whole Number Ratio: Calculate the smallest whole number ratio of C:H:Br.
4. Empirical Formula: Based on the ratio, derive the empirical formula (e.g., CH₂​Br).

Molecular Formula Calculation

1. Molar Mass of the Compound: Given molecular weight: 187.9 g/mol.
2. Molar Mass of the Empirical Formula: Calculate the molar mass based on the empirical formula.
3. Ratio of Molecular Weight to Empirical Formula Weight: Divide the given molecular weight by the empirical formula’s molar mass.
4. Determine the Molecular Formula: Multiply the empirical formula by the ratio obtained in the previous step.
5. Final Molecular Formula: The molecular formula of the compound (e.g., C₂​H₄​Br₂​).

Summary

The molecular formula of the compound is derived by first determining the empirical formula based on its percentage composition. The ratio of the molecular weight to the empirical formula weight gives the multiplier for the empirical formula to obtain the molecular formula. In this case, the molecular formula of the compound is determined to be C₂​H₄​Br₂​. This method is critical in the field of chemistry for identifying the actual molecular structure of compounds.