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Current Question (ID: 8575)

Question:
$\text{The balanced equation for the reaction between chlorine and sulphur dioxide in water is:}$
Options:
  • 1. \text{Cl}_2\text{(s)} + \text{SO}_2\text{(aq)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} + 4\text{H}^+\text{(aq)}
  • 2. 3\text{Cl}_2\text{(s)} + \text{SO}_2\text{(aq)} + 2\text{H}_2\text{O(l)} \rightarrow \text{Cl}^-\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} + 3\text{H}^+\text{(aq)}
  • 3. \text{Cl}_2\text{(s)} + 3\text{SO}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{Cl}^-\text{(aq)} + 2\text{SO}_4^{2-}\text{(aq)} + 4\text{H}^+\text{(aq)}
  • 4. 2\text{Cl}_2\text{(s)} + \text{SO}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} + 4\text{H}^+\text{(aq)}
Solution:
\text{Hint: Total charge should be balanced in the reaction.} \text{Explanation:} \text{The given redox reaction can be represented as:} \mathrm{Cl_2(s) + SO_2(aq) + H_2O(l) \rightarrow Cl^-(aq) + SO_4^{2-}(aq)} \text{The oxidation half reaction is:} \mathrm{S^{+4}O_2(aq) \rightarrow S^{+6}O_4^{2-}(aq)} \text{The oxidation number is balanced by adding two electrons as:} \mathrm{SO_2(aq) \rightarrow SO_4^{2-}(aq) + 2e^-} \text{The charge is balanced by adding 4H}^+ \text{ ions as:} \mathrm{SO_2(aq) \rightarrow SO_4^{2-}(aq) + 4H^+(aq) + 2e^-} \text{The O atoms and H}^+ \text{ ions are balanced by adding 2H}_2\text{O molecules as:} \mathrm{SO_2(aq) + 2H_2O(l) \rightarrow SO_4^{2-}(aq) + 4H^+(aq) + 2e^-} \quad \text{....(i)} \text{The reduction half reaction is:} \mathrm{Cl_2(s) \rightarrow Cl^-(aq)} \text{The chlorine atoms are balanced as:} \mathrm{Cl_2^0(s) \rightarrow 2Cl^{-1}(aq)} \text{The oxidation number is balanced by adding electrons:} \mathrm{Cl_2(s) + 2e^- \rightarrow 2Cl^-(aq)} \quad \text{......(ii)} \text{The balanced chemical equation can be obtained by adding equation (i) and (ii) as:} \mathrm{Cl_2(s) + SO_2(aq) + 2H_2O(l) \rightarrow 2Cl^-(aq) + SO_4^{2-}(aq) + 4H^+(aq)}

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{
  "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
}