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Current Question (ID: 10557)

Question:
$\text{A balanced chemical equation for the reaction showing catalytic oxidation of NH}_3 \text{ by atmospheric oxygen is:}$
Options:
  • 1. $4\text{NH}_3 + 5\text{O}_2 \xrightarrow[\text{500 K; 9 bar}]{\text{Pt / Rh gauge catalyst}} 4\text{NO} + 6\text{H}_2\text{O}$ (Correct)
  • 2. $3\text{NH}_4 + 5\text{O}_2 \xrightarrow[\text{500 K; 9 bar}]{\text{Pt / Rh gauge catalyst}} 3\text{NO} + 6\text{H}_2\text{O}$
  • 3. $4\text{NH}_3 + 5\text{O}_2 \xrightarrow[\text{cold}]{\text{no catalyst}} 4\text{NO} + 6\text{H}_2\text{O}$
  • 4. $3\text{NH}_4 + 5\text{O}_2 \xrightarrow[\text{cold}]{\text{no catalyst}} 3\text{NO} + 6\text{H}_2\text{O}$
Solution:
$\text{HINT: All the atoms should be balanced on both sides.}$ $\text{Explanation:}$ $\text{Ammonia (NH}_3\text{) on catalytic oxidation by atmospheric oxygen in presence of Rh/Pt gauge at 500K under the pressure of 9 bar produces nitrous oxide.}$ $\text{The balanced chemical reaction can be written as:}$ $4\text{NH}_3 + 5\text{O}_2 \xrightarrow[\text{500 K; 9 bar}]{\text{Pt / Rh gauge catalyst}} 4\text{NO} + 6\text{H}_2\text{O}$

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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
}