Import Question JSON

Current Question (ID: 9194)

Question:

$\text{The incorrect reaction among the following is:}$

Options:

1. $\text{Na}_2\text{O}_2\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{O}_2\text{(g)}$
2. $2\text{KO}_2\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{KOH(aq)} + \text{H}_2\text{O}_2\text{(aq)} + \text{O}_2\text{(g)}$
3. $\text{Na}_2\text{O(s)} + \text{CO}_2\text{(g)} \rightarrow \text{NaHCO}_3\text{(s)} + \text{H}_2\text{O}_2\text{(aq)}$
4. $\text{Na}_2\text{O(s)} + \text{CO}_2\text{(g)} \rightarrow \text{Na}_2\text{CO}_3\text{(s)}$

Solution:

$\text{Hint: Na}_2\text{O on reaction with CO}_2\text{ gives sodium carbonate.}$ $\text{Explanation:}$ $\text{1. The balanced chemical equation for the reaction between Na}_2\text{O}_2\text{ and water is:}$ $\text{Na}_2\text{O}_2\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{O}_2\text{(g)}$ $\text{2. The balanced chemical equation for the reaction between KO}_2\text{ and water is:}$ $2\text{KO}_2\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{KOH(aq)} + \text{H}_2\text{O}_2\text{(aq)} + \text{O}_2\text{(g)}$ $\text{4. The balanced chemical equation for the reaction between Na}_2\text{O and CO}_2\text{ is:}$ $\text{Na}_2\text{O(s)} + \text{CO}_2\text{(g)} \rightarrow \text{Na}_2\text{CO}_3\text{(s)}$ $\text{Reaction (3) is not correct.}$

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
}