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Current Question (ID: 7410)

Question:

$\text{A mixture of } 2.3 \text{ g formic acid and } 4.5 \text{ g oxalic acid is treated with conc. } \text{H}_2\text{SO}_4\text{. The evolved gaseous mixture is passed through KOH pellets.}\n\n\text{Weight (in g) of the remaining product at STP will be:}$

Options:

1. $1.4$
2. $3.0$
3. $2.8$
4. $4.4$

Solution:

$\text{Hint: KOH will absorb CO}_2 \text{ and conc. H}_2\text{SO}_4 \text{ will absorb water, so in final solution only CO will remain.}$ $\text{Explanation:}$ $\text{HCOOH} + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{O} + \text{CO}$ $\text{Initial } 0.05 \text{ mol} \quad \quad 0 \quad 0$ $\text{Final } 0 \quad \quad \quad 0.05 \text{ mol } 0.05 \text{ mol}$ $\text{H}_2\text{C}_2\text{O}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{O} + \text{CO} + \text{CO}_2$ $\text{Initial } 0.05 \text{ mol} \quad \quad 0 \quad 0 \quad 0$ $\text{Final } 0 \quad \quad \quad 0.05 \quad 0.05 \quad 0.05$ $\text{moles of HCOOH} = 2.3/46 = 0.05 \text{ mol}$ $\text{moles of oxalic acid} = 4.5/90 = 0.05 \text{ mol}$ $\text{KOH will absorb CO}_2 \text{ and conc. H}_2\text{SO}_4 \text{ will absorb water, so in final solution only CO will remain}$ $\text{Total moles of CO} = 0.05 \text{ mol} + 0.05 \text{ mol} = 0.1 \text{ mol}$ $\text{Mass of CO} = 0.1 \times 28 = 2.8 \text{ g CO}$

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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
}