Import Question JSON

Current Question (ID: 18821)

Question:

$\text{The values of X, Y, P, and Q are, respectively:}$ $X + \text{MnO}_2 + \text{O}_2$ $\Delta$ $\text{KMnO}_4$ $Y$ $\text{Explosive oil (P) if KMnO}_4 \text{ is in excess}$ $\text{Green (Q) if Y is in excess}$

Options:

1. $\text{MnO}_3^+, \text{Mn}_2\text{O}_7, \text{conc. H}_2\text{SO}_4, \text{and K}_2\text{MnO}_4$
2. $\text{K}_2\text{MnO}_4, \text{conc. H}_2\text{SO}_4, \text{Mn}_2\text{O}_7, \text{and MnO}_3^+$
3. $\text{Na}_2\text{CO}_3, \text{MnO}_2, \text{K}_2\text{MnO}_4, \text{and Mn}_2\text{O}_7$
4. $\text{H}^+, \text{K}_2\text{MnO}_4, \text{MnO}_3^+, \text{and Mn}_2\text{O}_7$

Solution:

$\text{Hint: Y is conc. H}_2\text{SO}_4$ $\text{The heating reaction of KMnO}_4 \text{ is as follows:}$ $\text{KMnO}_4 \xrightarrow{\Delta} \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2$ $\text{The X is K}_2\text{MnO}_4.$ $\text{When excess of KMnO}_4 \text{ reacts with conc. H}_2\text{SO}_4 \text{ then Mn}_2\text{O}_7 \text{ arises as a dark green oil. Mn}_2\text{O}_7 \text{ is highly explosive in nature. The reaction is as follows:}$ $2\text{KMnO}_4 + 2\text{H}_2\text{SO}_4 \rightarrow \text{Mn}_2\text{O}_7 + \text{H}_2\text{O} + 2\text{KHSO}_4$ $\text{Hence, Y is conc. H}_2\text{SO}_4 \text{ and P is Mn}_2\text{O}_7$ $\text{When KMnO}_4 \text{ reacts with excess of conc. H}_2\text{SO}_4 \text{ then Mn}_2\text{O}_7 \text{ is formed and Mn}_2\text{O}_7 \text{ further reacts with conc. H}_2\text{SO}_4 \text{ and [MnO}_3]^+ \text{ is formed as a product.}$ $\text{The reaction is as follows:}$ $\text{Mn}_2\text{O}_7 + 2\text{H}_2\text{SO}_4 \rightarrow 2[\text{MnO}_3]^+[\text{HSO}_4]^- + \text{H}_2\text{O}.$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}