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Current Question (ID: 8621)

Question:
$\text{The number of moles of MnO}_4^- \text{ required to oxidise one mole of ferrous oxalate completely in an acidic medium is-}$
Options:
  • 1. $0.6 \text{ mole}$
  • 2. $0.4 \text{ mole}$
  • 3. $7.5 \text{ moles}$
  • 4. $0.2 \text{ mole}$
Solution:
$10\text{FeC}_2\text{O}_4 + 6\text{KMnO}_4 + 24\text{H}_2\text{SO}_4 \rightarrow 3\text{K}_2\text{SO}_4 + 6\text{MnSO}_4 + 5\text{Fe}_2(\text{SO}_4)_3 + 24\text{H}_2\text{O} + 20\text{CO}_2$ $\text{So we see that 6 moles of KMnO}_4 \text{ is required to oxidize 10 moles of FeC}_2\text{O}_4$ $\text{Then, 1 mole of FeC}_2\text{O}_4 \text{ would be oxidized by } = 6/10 = 0.6$

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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
}