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Current Question (ID: 8595)

Question:
\text{The oxidizing agent and reducing agent in the given reaction are:} 3\text{N}_2\text{H}_4\text{(l)} + 4\text{ClO}_3^-\text{(aq)} \rightarrow 6\text{NO}\text{(g)} + 4\text{Cl}^-\text{(aq)} + 6\text{H}_2\text{O}\text{(l)}
Options:
  • 1. $\text{Oxidising agent = N}_2\text{H}_4\text{; Reducing agent = ClO}_3^-$
  • 2. $\text{Oxidising agent = ClO}_3^-\text{; Reducing agent = N}_2\text{H}_4$
  • 3. $\text{Oxidising agent = N}_2\text{H}_4\text{; Reducing agent = N}_2\text{H}_4$
  • 4. $\text{Oxidising agent = ClO}_3^-\text{; Reducing agent = ClO}_3^-$
Solution:
\text{Hint: An increase in oxidation state is oxidation, and a decrease in oxidation state is reduction.} \text{The reaction is as follows:} \text{The oxidation number of N increases from } -2 \text{ in } \text{N}_2\text{H}_4 \text{ to } +2 \text{ in NO and the oxidation number of Cl decreases from } +5 \text{ in } \text{ClO}_3^- \text{ to } -1 \text{ in } \text{Cl}^-. \text{Hence, in this reaction, } \text{N}_2\text{H}_4 \text{ is the reducing agent and } \text{ClO}_3^- \text{ is the oxidizing agent.}

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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
}