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Current Question (ID: 7399)

Question:
$\text{In the Haber process, 30 L of dihydrogen and 30 L of dinitrogen were taken for the reaction, which yielded only 50\% of the expected product. The composition of the gaseous mixture under the said conditions at the end is:}$
Options:
  • 1. $\text{20 L ammonia, 10 L nitrogen, 30 L hydrogen}$
  • 2. $\text{20 L ammonia, 25 L nitrogen, 15 L hydrogen}$
  • 3. $\text{20 L ammonia, 20 L nitrogen, 20 L hydrogen}$
  • 4. $\text{10 L ammonia, 25 L nitrogen, 15 L hydrogen}$
Solution:
$\text{Hint: Limiting reagent concept}$\n\n$\text{Step 1: First, let's write the balanced chemical equation for the Haber process:}$\n\n$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$\n\n$\text{According to the reaction stoichiometry:}$\n\n$1 \text{ volume of } \text{N}_2 \text{ reacts with } 3 \text{ volumes of } \text{H}_2 \text{ to produce } 2 \text{ volumes of } \text{NH}_3$\n\n$\text{Step 2: Determine the limiting reagent and the expected product}$\n\n$\text{For 30 L of } \text{N}_2\text{, the required } \text{H}_2 = 3 \times 30 = 90 \text{ L}$\n\n$\text{Since only 30 L of } \text{H}_2 \text{ is available, } \text{H}_2 \text{ is the limiting reagent.}$\n\n$\text{The maximum amount of } \text{N}_2 \text{ that can react with 30 L of } \text{H}_2 = \frac{30}{3} = 10 \text{ L}$\n\n$\text{The expected ammonia produced from 30 L of } \text{H}_2 = \frac{2 \times 30}{3} = 20 \text{ L}$\n\n$\text{Step 3: Calculate the actual product formed (given 50\% yield)}$\n\n$\text{Actual ammonia produced} = 50\% \text{ of } 20 \text{ L} = 10 \text{ L}$\n\n$\text{Step 4: Calculate the amounts of reactants consumed}$\n\n$\text{Since only 50\% of the expected reaction occurred, only 50\% of the limiting reagent (} \text{H}_2\text{) was consumed:}$\n\n$\text{H}_2 \text{ consumed} = 50\% \text{ of } 30 \text{ L} = 15 \text{ L}$\n\n$\text{N}_2 \text{ consumed} = \frac{\text{H}_2 \text{ consumed}}{3} = \frac{15}{3} = 5 \text{ L}$\n\n$\text{Step 5: Calculate the final composition of the gaseous mixture}$\n\n$\text{H}_2 \text{ remaining} = 30 - 15 = 15 \text{ L}$\n\n$\text{N}_2 \text{ remaining} = 30 - 5 = 25 \text{ L}$\n\n$\text{NH}_3 \text{ formed} = 10 \text{ L}$\n\n$\text{Therefore, the final composition of the gaseous mixture is 10 L ammonia, 25 L nitrogen, and 15 L hydrogen.}$

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{
  "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
}