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Current Question (ID: 18893)

Question:
$\text{The reaction(s) involved in the preparation of potassium dichromate from iron chromite ore is/are:}$ $1. \ \text{FeCr}_2\text{O}_4 + \text{NaOH} + \text{O}_2 \rightarrow \text{Na}_2\text{CrO}_4 + \text{Fe}_2\text{O}_3 + \text{H}_2\text{O}$ $2. \ \text{Na}_2\text{CrO}_4 + \text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$ $3. \ \text{Na}_2\text{Cr}_2\text{O}_7 + \text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + \text{NaCl}$ $4. \ \text{All of the above.}$
Options:
  • 1. $\text{FeCr}_2\text{O}_4 + \text{NaOH} + \text{O}_2 \rightarrow \text{Na}_2\text{CrO}_4 + \text{Fe}_2\text{O}_3 + \text{H}_2\text{O}$
  • 2. $\text{Na}_2\text{CrO}_4 + \text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$
  • 3. $\text{Na}_2\text{Cr}_2\text{O}_7 + \text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + \text{NaCl}$
  • 4. $\text{All of the above.}$
Solution:
$\text{Hint: The conversion occurs in three steps.}$ $\text{Step 1: Fusion of chromite ore (FeCr}_2\text{O}_4) \text{ with sodium or potassium carbonate in free access to air.}$ $4\text{FeCr}_2\text{O}_4 + 16\text{NaOH} + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{H}_2\text{O}$ $\text{Step 2: Conversion of sodium chromate into sodium dichromate.}$ $\text{Na}_2\text{CrO}_4 + 2\text{H}_2\text{SO}_4 \text{ (conc.)} \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$ $\text{Step 3: Conversion of sodium dichromate to potassium dichromate.}$ $\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}$ $\text{Potassium chloride is less soluble than sodium chloride is obtained in the form of orange colored crystals and can be removed by filtration.}$

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{
  "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
}