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Current Question (ID: 9148)

Question:

$\text{The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B. Excess of } \text{CO}_2 \text{ is bubbled through an aqueous solution of B. C is solid. Solid C on gentle heating gives back A. The compound A is:}$

Options:

1. $\text{Na}_2\text{CO}_3$
2. $\text{K}_2\text{CO}_3$
3. $\text{CaSO}_4 \cdot \text{2H}_2\text{O}$
4. $\text{CaCO}_3$

Solution:

$\text{HINT: } \text{CaCO}_3 \text{ on heating gives } \text{CO}_2 \text{ (colourless gas).}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{CaCO}_3 \text{ (A) on heating gives colourless gas(CO}_2\text{) and residues of CaO which in reaction with water gives Ca(OH)}_2 \text{ (B).}$ $\text{Step 2:}$ $\text{When excess } \text{CO}_2 \text{ is bubbled through an aqueous solution of Ca(OH)}_2 \text{ (B), it gives calcium bicarbonate Ca(HCO}_3\text{)}_2 \text{ (C) which on heating gives back CaCO}_3\text{(A).}$ $\text{The reactions are as follows:}$ $\text{CaCO}_3 \xrightarrow{\Delta} \text{CO}_2 + \text{CaO}$ $\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2$ $\text{Ca(OH)}_2 + \text{CO}_2\text{(excess)} \rightarrow \text{Ca(HCO}_3\text{)}_2$ $\text{Ca(HCO}_3\text{)}_2 \xrightarrow{\Delta} \text{CaCO}_3$

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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
}