Import Question JSON

Current Question (ID: 7391)

Question:

$\text{Assertion (A): In Haber's process, starting with 5 moles of } \text{N}_2 \text{ and 2.5 mol of } \text{H}_2\text{, on complete reaction only 1.66 moles of } \text{NH}_3 \text{ were produced.}$\n\n$\text{Reason (R): } \text{H}_2 \text{ acts as a limiting reagent in this reaction.}$

Options:

1. $\text{Both (A) and (R) are true and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are true but (R) is not the correct explanation of (A).}$
3. $\text{(A) is true but (R) is false.}$
4. $\text{Both (A) and (R) are false.}$

Solution:

$\text{Hint: Mole concepts}$\n\n$\text{The Haber process reaction is as follows:}$\n\n$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$\n\n$\text{As per the reaction, 1 mol of } \text{N}_2 \text{ reacts with 3 mol of } \text{H}_2\text{. So for 5 moles of } \text{N}_2 \text{, 15 moles of } \text{H}_2 \text{ is required.}$\n\n$\text{Hence, the amount of } \text{H}_2 \text{ given is 2.5 moles. } \text{H}_2 \text{ is a limiting reagent.}$\n\n$\text{As per the reaction, 3 moles of } \text{H}_2 \text{ gives 2 moles of } \text{NH}_3 \text{.}$\n\n$\text{1 mole of } \text{H}_2 \text{ gives } \frac{2}{3} \text{ moles of } \text{NH}_3$\n\n$\text{2.5 mol of } \text{H}_2 \text{ gives } \frac{2}{3} \times 2.5 \text{ moles of } \text{NH}_3$\n\n$= 1.66 \text{ moles of } \text{NH}_3$\n\n$\text{Hence, both Assertion \& Reason are true and the reason is the correct explanation of the assertion.}$

Import JSON File

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
}