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Current Question (ID: 8264)

Question:

$\text{During the complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for this change is:}$

Options:

1. 2\text{C}_4\text{H}_{10}\text{(g)} + 13\text{O}_2\text{(g)} \rightarrow 8\text{CO}_2\text{(g)} + 10\text{H}_2\text{O}\text{(l)}; \quad \Delta_c\text{H} = -2658.0 \text{ kJ mol}^{-1}
2. \text{C}_4\text{H}_{10}\text{(g)} + \frac{13}{2}\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 5\text{H}_2\text{O}\text{(l)}; \quad \Delta_c\text{H} = -1329.0 \text{ kJ mol}^{-1}
3. \text{C}_4\text{H}_{10}\text{(g)} + \frac{13}{2}\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 5\text{H}_2\text{O}\text{(l)}; \quad \Delta_c\text{H} = -2658.0 \text{ kJ mol}^{-1}
4. \text{C}_4\text{H}_{10}\text{(g)} + \frac{13}{2}\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 5\text{H}_2\text{O}\text{(l)}; \quad \Delta_c\text{H} = +2658.0 \text{ kJ mol}^{-1}

Solution:

$\text{HINT-Heat of combustion is heat released when 1 mole of any substance is burned in presence of oxygen}$ $\text{EXPLANATION}$ $\text{Given that, the complete combustion of one mole of butane is represented by thermochemical reaction as}$ $\text{C}_4\text{H}_{10}(g) + \frac{13}{2}\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 5\text{H}_2\text{O}(l)$ $\text{We have to take the combustion of one mole of C}_4\text{H}_{10} \text{ and } \Delta_c\text{H should be negative and have a value of 2658 kJ mol}^{-1}$ $\text{The answer is option third.}$

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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
}