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Current Question (ID: 18537)

Question:

$2 ext{MS} + 3 ext{O}_2 \rightarrow 2 ext{MO} + 2 ext{SO}_2$ $2 ext{MO} + ext{MS} \rightarrow 3 ext{M} + ext{SO}_2$ $\text{The metal M in above-mentioned reaction is:}$

Options:

1. $\text{Mg}$
2. $\text{Pb}$
3. $\text{Sn}$
4. $\text{Fe}$

Solution:

$\text{HINT: Pb is extracted by self reduction.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Self-reduction reaction of lead metal - In the self reduction process of lead sulphide is heated in air then lead sulphide is converted into lead oxide.}$ $\text{Lead oxide then reacts with the remaining lead sulphide to give lead metal. Thus self-reduction of lead is performed.}$ $\text{STEP 2:}$ $\text{The reactions involved are :}$ $2\text{PbS} + 3\text{O}_2 \rightarrow 2\text{PbO} + 2\text{SO}_2$ $2\text{PbO} + \text{PbS} \rightarrow 3\text{Pb} + \text{SO}_2$

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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
}