Import Question JSON

Current Question (ID: 7413)

Question:

$\begin{array}{|l|l|}\hline \text{Assertion (A):} & \text{When 4 moles of H}_2 \text{ reacts with 2 moles of O}_2\text{, then 4 moles of} \\ & \text{water are formed.} \\\hline \text{Reason (R):} & \text{O}_2 \text{ will act as a limiting reagent.} \\\hline\end{array}$

Options:

1. $\text{Both (A) and (R) are true and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are true but (R) is not the correct explanation of (A).}$
3. $\text{(A) is true but (R) is false.}$
4. $\text{Both (A) and (R) are false.}$

Solution:

$\text{Hint: The reaction is } 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ $\text{The reaction between H}_2 \text{ and O}_2 \text{ is as follows:}$ $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ $\text{From the reaction, it is clear that 2 moles H}_2 \text{ needs 1 mol O}_2\text{.}$ $1 \text{ mol H}_2 \text{ required, } \frac{1}{2} \text{ mol of O}_2$ $4 \text{ mole H}_2 \text{ required, } 4 \times \frac{1}{2} \text{ mol of O}_2$ $= 2 \text{ moles of O}_2 \text{ are required}$ $\text{Hence, O}_2 \text{ is not a limiting reagent.}$ $2 \text{ moles H}_2 \text{ produce 2 moles of water, thus, 4 moles of H}_2 \text{ produce 4 moles of water.}$ $\text{Thus, Assertion is a true statement but Reason is false.}$

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
}