Import Question JSON

Current Question (ID: 8572)

Question:

\text{In the given balanced chemical reaction: } \text{IO}_3^- + a\text{I}^- + b\text{H}^+ \rightarrow c\text{H}_2\text{O} + d\text{I}_2 \text{. The values of a, b, c, and d respectively are:}

Options:

1. $5, 6, 3, 3$
2. $5, 3, 6, 3$
3. $3, 5, 3, 6$
4. $5, 6, 5, 5$

Solution:

$\text{Step 1: Identify oxidation and reduction half reactions}$ $\text{Reduction: } \text{IO}_3^- \rightarrow \text{I}_2^0$ $\text{Oxidation: } \text{I}^- \rightarrow \text{I}_2$ $\text{Step 2: Balance them atom-wise}$ $2\text{IO}_3^- \rightarrow \text{I}_2$ $2\text{I}^- \rightarrow \text{I}_2$ $\text{Step 3: Balance them charge-wise}$ $2\text{IO}_3^- + 10\text{e} + 12\text{H}^+ \rightarrow \text{I}_2^0 + 6\text{H}_2\text{O}$ $2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}$ $\text{Step 4: Balance electrons by multiplying oxidation half-reaction by 5}$ $(2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}) \times 5$ $\text{Step 5: Add the balanced half-reactions}$ $2\text{IO}_3^- + 12\text{H}^+ + 10\text{I}^- \rightarrow 6\text{I}_2 + 6\text{H}_2\text{O}$ $\text{Step 6: Simplify by dividing by 2}$ $\text{IO}_3^- + 6\text{H}^+ + 5\text{I}^- \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}$ $\text{Therefore: a = 5, b = 6, c = 3, d = 3}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}