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Current Question (ID: 8638)

Question:

$\text{In the reaction:}$ $\text{Al} + \text{Fe}_3\text{O}_4 \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}$ $\text{How many electrons are transferred in total during the reaction?}$

Options:

1. $6$
2. $8$
3. $\frac{8}{3}$
4. $24$

Solution:

$\text{Hint: The oxidation state of Al is changed from 0 to +3}$ $\text{The two half cell reactions are as follows:}$ $\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 6\text{e}^- \quad \ldots(1)$ $8\text{e}^- + \text{Fe}_3\text{O}_4 \rightarrow \text{Fe} \quad \ldots(2)$ $\text{Multiplying equation (1) with 4 and equation (2) with 3}$ $4 \times \text{Al} \rightarrow \text{Al}_2\text{O}_3 + 6\text{e}^- \quad \ldots(1)$ $3 \times 8\text{e}^- + \text{Fe}_3\text{O}_4 \rightarrow \text{Fe} \quad \ldots(2)$ $4\text{Al} \rightarrow 2\text{Al}_2\text{O}_3 + 24\text{e}^- \quad \ldots(1)$ $24\text{e}^- + \text{Fe}_3\text{O}_4 \rightarrow 3\text{Fe} \quad \ldots(2)$ $\text{Now balance the number of electrons. Multiplying equation (1) with 4 and equation (2) with 3 total e}^- \text{ transferred} = 24$

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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
}