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Current Question (ID: 10559)

Question:
$\text{Nitrogen gas is liberated by the thermal decomposition of:}$
Options:
  • 1. $\text{NH}_4\text{NO}_2$
  • 2. $\text{NaN}_3$
  • 3. $(\text{NH}_4)_2\text{Cr}_2\text{O}_7$
  • 4. $\text{All of the above}$ (Correct)
Solution:
$\text{Hint: Pure nitrogen is obtained in small amounts by heating sodium or barium azides in a vacuum.}$ $\text{Explanation:}$ $\text{The nitrogen synthesis reactions are as follows:}$ $\text{a. Nitrogen in the laboratory can be obtained by heating ammonium nitrite or ammonium dichromate. Ammonium nitrite is not a stable compound and thus it is first formed as an intermediate product by heating an ammonium salt with sodium nitrite.}$ $\text{The reaction is as follows:}$ $\text{NH}_4\text{Cl} + \text{NaNO}_2 \rightarrow \text{NH}_4\text{NO}_2 + \text{NaCl}$ $\text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}$ $(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \rightarrow \text{N}_2 + \text{Cr}_2\text{O}_3 + 4\text{H}_2\text{O}$ $\text{b. Pure nitrogen is obtained in small amounts by heating sodium or barium azides in a vacuum.}$ $\text{NaN}_3 \xrightarrow{\Delta} \text{Na} + \frac{3}{2}\text{N}_2$ $\text{Hence, option fourth is the correct answer.}$

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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
}