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Current Question (ID: 20844)

Question:

$\text{Consider the reaction}$ $4\text{HNO}_3(\text{l}) + 3\text{KCl}(\text{s}) \rightarrow \text{Cl}_2(\text{g}) + \text{NOCl}(\text{g}) + 2\text{H}_2\text{O}(\text{g}) + 3\text{KNO}_3(\text{s})$ $\text{The amount of HNO}_3 \text{ required to produce 110.0 g of KNO}_3 \text{ is:}$ $\text{(Given: Atomic masses of H, O, N and K are 1, 16, 14 and 39 respectively).}$

Options:

1. $32.2 \text{ g}$
2. $69.4 \text{ g}$
3. $91.5 \text{ g}$
4. $162.5 \text{ g}$

Solution:

$\text{Hint: Use mole concept}$ $4\text{HNO}_3(\text{l}) + 3\text{KCl}(\text{s}) \rightarrow \text{Cl}_2(\text{g}) + \text{NOCl}(\text{g}) + 2\text{H}_2\text{O}(\text{g}) + 3\text{KNO}_3(\text{s})$ $\therefore \text{ 110 g of KNO}_3 \Rightarrow \text{ moles of KNO}_3 = \frac{110}{101} = 1.089 \text{ mol}$ $\text{As, 4 mole of HNO}_3 \text{ produces 3 mol of KNO}_3. \text{ Hence, the moles of HNO}_3 \text{ required to produce 1.089 moles of KNO}_3 = \frac{4}{3} \times 1.089 = 1.452 \text{ mol}$ $\text{Hence, mass of HNO}_3 \text{ required is } 1.452 \times 63 = 91.5 \text{ g (approx.)}$

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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
}