Import Question JSON

Current Question (ID: 8374)

Question:

\text{For a reaction, } 2\text{NO}_{(g)} + \text{Br}_{2(g)} \rightleftharpoons 2\text{NOBr}_{(g)} \text{When 0.087 mol of NO and 0.0437 mol of Br}_2 \text{ are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. The concentration of NO and Br}_2 \text{ at equilibrium will be:}

Options:

1. $\text{NO} = 0.0352 \text{ mol}; \text{Br}_2 = 0.0178 \text{ mol}$
2. $\text{NO} = 0.352 \text{ mol}; \text{Br}_2 = 0.178 \text{ mol}$
3. $\text{NO} = 0.0634 \text{ mol}; \text{Br}_2 = 0.0596 \text{ mol}$
4. $\text{NO} = 0.634 \text{ mol}; \text{Br}_2 = 0.596 \text{ mol}$

Solution:

\text{Hint: Use stoichiometric calculations to solve this question.} \text{Explanation:} \text{Step 1: Write down the equilibrium conditions} \text{The given reaction is:} 2\text{NO}_{(g)} + \text{Br}_{2(g)} \rightleftharpoons 2\text{NOBr}_{(g)} \text{Stoichiometry: } 2 \text{ mol} \quad 1 \text{ mol} \quad 2 \text{ mol} \text{From stoichiometry: 2 mol of NOBr is formed from 2 mol of NO.} \text{Therefore, 0.0518 mol of NOBr is formed from 0.0518 mol of NO.} \text{Also, 2 mol of NOBr is formed from 1 mol of Br}_2\text{.} \text{Therefore, 0.0518 mol of NOBr is formed from } \frac{0.0518}{2} \text{ mol of Br}_2\text{.} \text{This equals 0.0259 mol of Br}_2\text{.} \text{Step 2: Calculate the amount of NO and Br}_2 \text{Initial amounts:} \text{NO} = 0.087 \text{ mol}; \quad \text{Br}_2 = 0.0437 \text{ mol} \text{Amount of NO at equilibrium:} \text{NO} = 0.087 - 0.0518 = 0.0352 \text{ mol} \text{Amount of Br}_2 \text{ at equilibrium:} \text{Br}_2 = 0.0437 - 0.0259 = 0.0178 \text{ mol}

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
}