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Current Question (ID: 20817)

Question:
$\text{Consider the following combustion reaction:}$ $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ $\text{How many moles of methane (CH}_4\text{) will be required for the formation of 11 g of carbon dioxide (CO}_2\text{)?}$
Options:
  • 1. 0.25
  • 2. 0.62
  • 3. 0.17
  • 4. 1.23
Solution:
$\text{Hint: Moles of CO}_2 = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2}$ $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ $1 \text{ mole of CH}_4 \text{ will produce 1 mole of CO}_2$ $\text{So, 11 g of CO}_2 \text{ will be produced by } \frac{11}{44} \text{ moles of CH}_4$ $\text{i.e., } \frac{1}{4} \text{ moles of CH}_4 = 0.25$

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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
}