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Current Question (ID: 8581)

Question:

$\text{In an alkaline medium } \text{ClO}_2 \text{ oxidize } \text{H}_2\text{O}_2 \text{ in } \text{O}_2 \text{ and reduces itself in } \text{Cl}^-. \text{ How many moles of } \text{H}_2\text{O}_2 \text{ will be oxidized by one mole of } \text{ClO}_2?$

Options:

1. $1.0$
2. $1.5$
3. $2.5$
4. $3.5$

Solution:

$\text{Hint = Balance the equation.}$ $\text{Reduction half-reaction: } \text{ClO}_2 \rightarrow \text{Cl}^-$ $\text{ClO}_2 + 2\text{H}_2\text{O}_2 + 5\text{e}^- \rightarrow \text{Cl}^- + 4\text{OH}^-$ $\text{Oxidation half-reaction: } \text{H}_2\text{O}_2 \rightarrow \text{O}_2$ $\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2\text{e}^-$ $\text{Balance electrons by multiplying:}$ $[\text{ClO}_2 + 2\text{H}_2\text{O}_2 + 5\text{e}^- \rightarrow \text{Cl}^- + 4\text{OH}^-] \times 2$ $[\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2\text{e}^-] \times 5$ $\text{Overall balanced equation:}$ $2\text{ClO}_2 + 5\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow 2\text{Cl}^- + 5\text{O}_2 + 6\text{H}_2\text{O}$ $\text{Stoichiometric ratio: } 2\text{ClO}_2 \equiv 5\text{H}_2\text{O}_2$ $\therefore \text{ClO}_2 = 2.5 \text{H}_2\text{O}_2$

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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
}