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Current Question (ID: 18844)

Question:

$\text{When iron becomes passive by "X", it is due to the formation of "Y". X and Y, respectively, are:}$

Options:

1. $\text{Dil. HCl, Fe}_2\text{O}_3$
2. $80\% \text{ Conc. HNO}_3, \text{ Fe}_3\text{O}_4$
3. $\text{Conc. H}_2\text{SO}_4, \text{Fe}_3\text{O}_4$
4. $\text{Conc. HCl, Fe}_3\text{O}_4$

Solution:

$\text{HINT: Iron becomes passive by conc. nitric acid.}$ $\text{Explanation:}$ $\text{STEP 1: Metals like iron, aluminium, cobalt, and nickel become inert or passive when exposed to pure concentrated nitric acid due to the formation of a thin layer of insoluble metallic oxide on the surface.}$ $\text{This newly formed layer prevents the bulk from reacting any further. Because concentrated nitric acid is a strong oxidizer, it transforms the metal to ferric oxide, which forms an insoluble protective covering.}$ $\text{STEP 2: When the metal loses contact with the acid, the process comes to an end. When a chemically active metal becomes non-reactive and inactive, it loses its chemical activity.}$ $\text{Thus, Iron becomes passive with fuming nitric acid due to the formation of a reddish-brown covering of iron oxide (Fe}_3\text{O}_4) \text{ that prevents it from reacting.}$ $2\text{HNO}_3 \rightarrow \text{H}_2\text{O} + 2\text{NO}_2 + [\text{O}]$ $3\text{Fe} + 4[\text{O}] \rightarrow \text{Fe}_3\text{O}_4$ $3\text{Fe} + 8\text{HNO}_3 \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2\text{O} + 8\text{NO}_2$ $\text{(Similar behavior is observed in case of Al, Cr, Co and Ni also.)}$

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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
}