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Current Question (ID: 7407)

Question:
$\text{In the reaction, } 4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 4\text{NO(g)} + 6\text{H}_2\text{O(l)}$\n\n$\text{When 1 mole of ammonia and 1 mole of } \text{O}_2 \text{ reacts to completion, then:}$
Options:
  • 1. $\text{1.0 mole of } \text{H}_2\text{O is produced.}$
  • 2. $\text{1.0 mole of NO will be produced.}$
  • 3. $\text{All the oxygen will be consumed.}$
  • 4. $\text{All the ammonia will be consumed.}$
Solution:
$\text{Hint: Mole concept}$\n\n$\text{Step 1: Analyze the balanced chemical equation and the stoichiometric ratios.}$\n\n$4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 4\text{NO(g)} + 6\text{H}_2\text{O(l)}$\n\n$\text{From the balanced equation, the molar ratios are:}$\n\n$4 \text{ moles of } \text{NH}_3 : 5 \text{ moles of } \text{O}_2 : 4 \text{ moles of NO} : 6 \text{ moles of } \text{H}_2\text{O}$\n\n$\text{Step 2: Determine which reactant is the limiting reagent.}$\n\n$\text{For } \text{NH}_3\text{:}$\n\n$1 \text{ mole of } \text{NH}_3 \text{ requires } \frac{5}{4} = 1.25 \text{ moles of } \text{O}_2$\n\n$\text{But we only have 1 mole of } \text{O}_2 \text{ available.}$\n\n$\text{For } \text{O}_2\text{:}$\n\n$1 \text{ mole of } \text{O}_2 \text{ requires } \frac{4}{5} = 0.8 \text{ moles of } \text{NH}_3$\n\n$\text{We have 1 mole of } \text{NH}_3 \text{ available, which is more than needed.}$\n\n$\text{Since we need 1.25 moles of } \text{O}_2 \text{ but only have 1 mole, } \text{O}_2 \text{ is the limiting reagent.}$\n\n$\text{Step 3: Calculate the amounts of reactants consumed and products formed.}$\n\n$\text{Since } \text{O}_2 \text{ is the limiting reagent, all of the oxygen will be consumed.}$\n\n$\text{Amount of } \text{NH}_3 \text{ consumed} = 1 \text{ mole of } \text{O}_2 \times \frac{4 \text{ moles } \text{NH}_3}{5 \text{ moles } \text{O}_2} = 0.8 \text{ moles of } \text{NH}_3$\n\n$\text{This means that not all of the ammonia will be consumed; } 1 - 0.8 = 0.2 \text{ moles of } \text{NH}_3 \text{ will remain.}$\n\n$\text{Amount of NO produced} = 1 \text{ mole of } \text{O}_2 \times \frac{4 \text{ moles NO}}{5 \text{ moles } \text{O}_2} = 0.8 \text{ moles of NO}$\n\n$\text{Amount of } \text{H}_2\text{O produced} = 1 \text{ mole of } \text{O}_2 \times \frac{6 \text{ moles } \text{H}_2\text{O}}{5 \text{ moles } \text{O}_2} = 1.2 \text{ moles of } \text{H}_2\text{O}$\n\n$\text{Therefore, all the oxygen will be consumed, which corresponds to option 3.}$

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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
}