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Current Question (ID: 7394)

Question:

$\text{The mass of ammonia produced when } 2.00 \times 10^3 \text{ g dinitrogen reacts with } 1.00 \times 10^3 \text{ g of dihydrogen is:}$

Options:

1. $2338.11 \text{ g}$
2. $2428.57 \text{ g}$
3. $2712.24 \text{ g}$
4. $2180.56 \text{ g}$

Solution:

$\text{Hint: Use Limiting reagent concept on the balanced equation.}$\n\n$\text{Explanation:}$\n\n$\text{Reaction: } \text{N}_2 \text{ (g) } + 3\text{H}_2 \text{ (g) } \rightarrow 2\text{NH}_3 \text{ (g)}$\n\n$\text{Let's analyze the molar relationships in this reaction:}$\n\n$\text{For 1 mole of } \text{N}_2 \text{: Molar mass = 28 g}$\n\n$\text{For 3 moles of } \text{H}_2 \text{: Molar mass = 6 g}$\n\n$\text{For 2 moles of } \text{NH}_3 \text{: Molar mass = 34 g}$\n\n$\text{First, let's determine the required mass of } \text{H}_2 \text{ to completely react with } 2.00 \times 10^3 \text{ g of } \text{N}_2\text{:}$\n\n$\text{Required mass of } \text{H}_2 = \frac{6\text{ g}}{28\text{ g}} \times (2 \times 10^3 \text{ g}) = 428.6 \text{ g of } \text{H}_2$\n\n$\text{Given, the amount of } \text{H}_2 = 1.00 \times 10^3 \text{ g, which is more than 428.6 g of } \text{H}_2$\n\n$\text{Since we have more } \text{H}_2 \text{ than required, } \text{N}_2 \text{ is the limiting reagent.}$\n\n$\text{Thus, the mass of } \text{NH}_3 \text{ produced by } 2 \times 10^3 \text{ g of } \text{N}_2 \text{ can be calculated:}$\n\n$\text{Mass of } \text{NH}_3 = \frac{34\text{ g}}{28\text{ g}} \times (2 \times 10^3 \text{ g}) = 2428.57 \text{ g}$

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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
}