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Current Question (ID: 20863)

Question:

$\text{Consider the above reaction, the limiting reagent of the reaction and number of moles of NH}_3 \text{ formed respectively are:}$

Options:

1. $\text{H}_2, 1.42 \text{ moles}$
2. $\text{H}_2, 0.71 \text{ moles}$
3. $\text{N}_2, 1.42 \text{ moles}$
4. $\text{N}_2, 0.71 \text{ moles}$

Solution:

$\text{Hint: Limiting reagent: The reactant that gets consumed first in a chemical reaction and limits the amount of product formed.}$ $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ $\text{Ideally 28 g N}_2 \text{ reacts with 6 g H}_2 \text{ limiting reagent is N}_2$ $\text{Amount of NH}_3 \text{ formed on reacting 20 g N}_2 \text{ is,}$ $= \frac{34 \times 20}{28} = 24.28 \text{ g}$ $= 1.42 \text{ moles}$

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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
}