Import Question JSON

Current Question (ID: 18484)

Question:

$\text{Reactions among the following that occur during calcination are:}$ $\begin{array}{l} \text{a. } \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \\ \text{b. } 2\text{FeS}_2 + \frac{11}{2} \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2 \\ \text{c. } \text{Al}_2\text{O}_3\cdot\text{xH}_2\text{O} \rightarrow \text{Al}_2\text{O}_3 + \text{xH}_2\text{O} \\ \text{d. } \text{ZnS} + \frac{3}{2} \text{O}_2 \rightarrow \text{ZnO} + \text{SO}_2 \end{array}$ $\text{Choose the correct option:}$ $\text{1. a and b}$ $\text{2. b and c}$ $\text{3. c and d}$ $\text{4. a and c}$

Options:

1. $\text{a and b}$
2. $\text{b and c}$
3. $\text{c and d}$
4. $\text{a and c}$

Solution:

$\text{HINT: Calcination = Absence of air or limited supply of air.}$ $\text{Explanation:}$ $\text{Calcination involves heating of the ore below its melting point in the absence}$ $\text{of air or in limited supply of air. Oxygen containing ores like oxide, hydroxides}$ $\text{and carbonates are calcined.}$ $\text{Thus, the following reactions occur during calcination.}$ $\text{CaCO}_3 \overset{\Delta}{\rightarrow} \text{CaO} + \text{CO}_2$ $\text{Al}_2\text{O}_3\cdot\text{xH}_2\text{O} \overset{\Delta}{\rightarrow} \text{Al}_2\text{O}_3 + \text{xH}_2\text{O}$

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
}