Import Question JSON

Current Question (ID: 9207)

Question:

$\text{The byproduct of the Solvay process is:}$

Options:

1. $\text{CO}_2$
2. $\text{NH}_3$
3. $\text{CaCl}_2$
4. $\text{CaCO}_3$

Solution:

$\text{HINT: } \text{CaCl}_2 \text{ is byproduct of Solvay's process.}$ $\text{Explanation:}$ $\text{STEP 1: In Solvay process, also known as ammonia soda process, carbon dioxide is passed through a brine solution (containing about 28\% NaCl) which is saturated with ammonia to form sodium carbonate.}$ $\text{The reaction is as follows:}$ $2\text{NH}_3 + \text{H}_2\text{O} + \text{CO}_2 \rightarrow (\text{NH}_4)_2\text{CO}_3$ $(\text{NH}_4)_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \rightarrow 2\text{NH}_4\text{HCO}_3$ $\text{NH}_4\text{HCO}_3 + \text{NaCl} \rightarrow \text{NaHCO}_3 \downarrow + \text{NH}_4\text{Cl}$ $\text{STEP 2: The precipitate of sodium bicarbonate is filtered, dried and ignited to form sodium carbonate.}$ $2\text{NaHCO}_3 \text{ upon heating gives } \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$ $\text{STEP 3:}$ $\text{The carbon dioxide required for the reaction can be obtained by heating limestone (calcium carbonate) to 1300 K in a lime kiln.}$ $\text{Lime dissolves in water to form calcium hydroxide which is then transferred to the ammonia recovery tower.}$ $\text{CaCO}_3 \xrightarrow{\text{heat}} \text{CaO} + \text{CO}_2$ $\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2$ $\text{STEP 4: Ammonia required for the process can be prepared by heating ammonium chloride with calcium hydroxide.}$ $\text{Ca(OH)}_2 + 2\text{NH}_4\text{Cl} \rightarrow \text{CaCl}_2 + 2\text{NH}_3 + 2\text{H}_2\text{O}$

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
}