Import Question JSON

Current Question (ID: 7393)

Question:

$\text{In a closed vessel, 50 ml of } \text{A}_2\text{B}_3 \text{ completely reacts with 200 ml of } \text{C}_2 \text{ according to the following equation:}$\n\n$2\text{A}_2\text{B}_3\text{(g)} + 5\text{C}_2\text{(g)} \rightarrow 3\text{C}_3\text{B}_2\text{(g)} + \text{CA}_4\text{(g)}$\n\n$\text{The composition of the gaseous mixture in the system will be:}$

Options:

1. $100 \text{ ml } \text{C}_2\text{, } 50 \text{ ml } \text{C}_3\text{B}_2\text{, } 50 \text{ ml } \text{CA}_4$
2. $25 \text{ ml } \text{C}_2\text{, } 75 \text{ ml } \text{C}_3\text{B}_2\text{, } 25 \text{ ml } \text{CA}_4$
3. $75 \text{ ml } \text{C}_2\text{, } 75 \text{ ml } \text{C}_3\text{B}_2\text{, } 25 \text{ ml } \text{CA}_4$
4. $10 \text{ ml } \text{C}_2\text{, } 25 \text{ ml } \text{C}_3\text{B}_2\text{, } 100 \text{ ml } \text{CA}_4$

Solution:

$\text{Hint: Apply the concept of limiting reagent}$\n\n$2\text{A}_2\text{B}_3 + 5\text{C}_2 \rightarrow 3\text{C}_3\text{B}_2 + \text{CA}_4$\n\n$\text{Initial volumes: } 50 \text{ ml of } \text{A}_2\text{B}_3 \text{ and } 200 \text{ ml of } \text{C}_2$\n\n$\text{According to the balanced equation, for 2 volumes of } \text{A}_2\text{B}_3\text{, } 5 \text{ volumes of } \text{C}_2 \text{ are required.}$\n\n$\text{So for } 50 \text{ ml of } \text{A}_2\text{B}_3\text{, the required } \text{C}_2 \text{ should be:}$\n\n$\text{Required } \text{C}_2 = \frac{5}{2} \times 50 = 125 \text{ ml}$\n\n$\text{Since 200 ml of } \text{C}_2 \text{ is provided, which is more than required, } \text{A}_2\text{B}_3 \text{ will act as the limiting reagent.}$\n\n$\text{Unreacted } \text{C}_2 \text{ will be } 200 - 125 = 75 \text{ ml}$\n\n$\text{From the balanced equation:}$\n\n$\text{For 2 volumes of } \text{A}_2\text{B}_3\text{, } 3 \text{ volumes of } \text{C}_3\text{B}_2 \text{ are produced.}$\n\n$\text{So for } 50 \text{ ml of } \text{A}_2\text{B}_3\text{, the amount of } \text{C}_3\text{B}_2 \text{ produced will be:}$\n\n$\text{Volume of } \text{C}_3\text{B}_2 = \frac{3}{2} \times 50 = 75 \text{ ml}$\n\n$\text{Similarly, for 2 volumes of } \text{A}_2\text{B}_3\text{, } 1 \text{ volume of } \text{CA}_4 \text{ is produced.}$\n\n$\text{So for } 50 \text{ ml of } \text{A}_2\text{B}_3\text{, the amount of } \text{CA}_4 \text{ produced will be:}$\n\n$\text{Volume of } \text{CA}_4 = \frac{1}{2} \times 50 = 25 \text{ ml}$\n\n$\text{Therefore, the composition of the gaseous mixture in the system will be:}$\n\n$75 \text{ ml of } \text{C}_2\text{, } 75 \text{ ml of } \text{C}_3\text{B}_2\text{, and } 25 \text{ ml of } \text{CA}_4$

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
}