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Current Question (ID: 20805)

Question:

$ ext{C}(s) + ext{O}_2(g) ightarrow ext{CO}_2(g) + 400 ext{kJ}$ $ ext{C}(s) + \frac{1}{2} ext{O}_2(g) ightarrow ext{CO}(g) + 100 ext{kJ}$ $\text{When coal of purity } 60\% \text{ is allowed to burn in presence of insufficient oxygen, } 60\% \text{ of carbon is converted into 'CO' and the remaining is converted into 'CO}_2\text{'. The heat generated when } 0.6 \text{ kg of coal is burnt is:}$

Options:

1. $1600 \text{kJ}$
2. $3200 \text{kJ}$
3. $4400 \text{kJ}$
4. $6600 \text{kJ}$

Solution:

$\text{Mass of carbon} = (0.6 \times 10^3) \frac{60}{100} = \frac{600 \times 60}{100} = 360 \text{ gram}$ $60\% \text{ of carbon} \Rightarrow \frac{360 \times 60}{100} = 216 \text{ gram}$ $(1) \text{C}(s) + \frac{1}{2} \text{O}_2 \rightarrow \text{CO}(g) : \Delta H^\circ = -100 \text{kJ/mole}$ $\left(\frac{216}{12}\right)$ $\Delta H = (-100) \frac{216}{12}$ $= -1800 \text{kJ}$ $(2) \text{C}(s) + \text{O}_2 \rightarrow \text{CO}_2(g) : \Delta H^\circ = -400 \text{kJ/mole}$ $\left(\frac{144}{12}\right)$ $\Delta H = (-400) \times \frac{144}{12} = -4800 \text{kJ}$ $\text{Total heat released} = (1800 + 4800) = 6600 \text{kJ}$

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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
}