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Current Question (ID: 20852)

Question:

$\text{NaClO}_3 \text{ is used, even in spacecrafts, to produce } \text{O}_2 . \text{ The daily consumption of pure } \text{O}_2 \text{ by a person is } 492 \text{ L at } 1 \text{ atm, } 300 \text{ K. How much amount of NaClO}_3 , \text{ in grams, is required to produce } \text{O}_2 \text{ for the daily consumption of a person at } 1 \text{ atm, } 300 \text{ K?}$ $\text{NaClO}_3(\text{s}) + \text{Fe(s)} \rightarrow \text{O}_2(\text{g}) + \text{NaCl(s)} + \text{FeO(s)}$ $R = 0.082 \text{ L atm mol}^{-1} \text{ K}^{-1}$

Options:

1. $2130 \text{ gm}$
2. $106.5 \text{ gm}$
3. $4123 \text{ gm}$
4. $2912 \text{ gm}$

Solution:

$\text{Hint: Apply mole concept}$ $\text{NaClO}_3(\text{s}) + \text{Fe(s)} \rightarrow \text{O}_2(\text{g}) + \text{NaCl(s)} \quad 492 \text{ lit}$ $n_{\text{O}_2} = \frac{1 \times 492}{0.082 \times 300}$ $n_{\text{O}_2} = \frac{4920}{82 \times 3} = 20 \text{ mol}$ $\text{moles of NaClO}_3 = 20 \text{ mol}$ $\text{mass of NaClO}_3 = 20 \times 106.5 = 2130 \text{ gm}$

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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
}