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Current Question (ID: 7395)

Question:

$\text{If 0.5 moles of } \text{BaCl}_2 \text{ is reacted with 0.2 moles of } \text{Na}_3\text{PO}_4 \text{ then the maximum moles of } \text{Ba}_3\text{(PO}_4\text{)}_2 \text{ formed is:}$

Options:

1. $0.33$
2. $0.25$
3. $0.10$
4. $0.52$

Solution:

$\text{Hint: Use mole concept and limiting reagent concept.}$\n\n$\text{Step 1:}$\n\n$\text{Let's write the balanced chemical equation:}$\n\n$3\text{BaCl}_2 + 2\text{Na}_3\text{PO}_4 \rightarrow \text{Ba}_3\text{(PO}_4\text{)}_2 + 6\text{NaCl}$\n\n$\text{From the balanced equation:}$\n\n$3 \text{ moles of } \text{BaCl}_2 \text{ need } 2 \text{ moles of } \text{Na}_3\text{PO}_4$\n\n$1 \text{ mole of } \text{BaCl}_2 \text{ needs } \frac{2}{3} \text{ moles of } \text{Na}_3\text{PO}_4$\n\n$0.5 \text{ moles of } \text{BaCl}_2 \text{ needs } \frac{2}{3} \times 0.5 \text{ moles of } \text{Na}_3\text{PO}_4$\n\n$= 0.33 \text{ moles of } \text{Na}_3\text{PO}_4$\n\n$\text{The amount of } \text{Na}_3\text{PO}_4 \text{ required is } 0.33 \text{ moles, but the given amount is } 0.2 \text{ moles.}$\n\n$\text{Therefore, } \text{Na}_3\text{PO}_4 \text{ is the limiting reagent.}$\n\n$\text{Step 2:}$\n\n$\text{From the balanced equation:}$\n\n$2 \text{ moles of } \text{Na}_3\text{PO}_4 \text{ produces } 1 \text{ mole of } \text{Ba}_3\text{(PO}_4\text{)}_2$\n\n$0.2 \text{ moles of } \text{Na}_3\text{PO}_4 \text{ will produce } \frac{1}{2} \times 0.2 \text{ moles of } \text{Ba}_3\text{(PO}_4\text{)}_2$\n\n$= 0.1 \text{ moles of } \text{Ba}_3\text{(PO}_4\text{)}_2$\n\n$\text{Therefore, the maximum moles of } \text{Ba}_3\text{(PO}_4\text{)}_2 \text{ formed is } 0.1 \text{ moles.}$

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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
}