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Current Question (ID: 8576)

Question:

\text{For the redox reaction,} a\text{MnO}_4^- + b\text{C}_2\text{O}_4^{2-} + c\text{H}^+ \rightarrow d\text{Mn}^{2+} + e\text{CO}_2 + f\text{H}_2\text{O} \text{the correct stoichiometric coefficients of the reactants a, b, and c respectively for the balanced equation are:}

Options:

1. $ 16, 5, 2 $
2. $ 2, 5, 16 $
3. $ 2, 16, 5 $
4. $ 5, 16, 2 $

Solution:

\text{According to ion-electron method:} \text{Oxidation half reaction:} \quad \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^- \text{Multiply by 5} \text{Reduction half reaction:} \quad \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \text{Multiply by 2} \text{Combining the two half-reactions after multiplying to balance electrons:} 5(\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-) \Rightarrow 5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10\text{e}^- 2(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}) \Rightarrow 2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \text{Adding the two balanced half-reactions:} 5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 10\text{CO}_2 + 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \text{Comparing this with the given general reaction:} a\text{MnO}_4^- + b\text{C}_2\text{O}_4^{2-} + c\text{H}^+ \rightarrow d\text{Mn}^{2+} + e\text{CO}_2 + f\text{H}_2\text{O} \text{We find the coefficients of the reactants:} a = 2, \quad b = 5, \quad c = 16 \text{Therefore, the correct stoichiometric coefficients are a = 2, b = 5, and c = 16.}

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}