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Current Question (ID: 7405)

Question:
$\text{25.4 g of } \text{I}_2 \text{ and 14.2 g of } \text{Cl}_2 \text{ are made to react completely to yield a mixture of ICl and } \text{ICl}_3\text{. The mole of ICl and } \text{ICl}_3 \text{ formed, is respectively -}$
Options:
  • 1. $0.5, 0.2$
  • 2. $0.1, 0.1$
  • 3. $0.1, 0.3$
  • 4. $0.3, 0.4$
Solution:
$\text{Hint: Mole concept}$\n\n$\text{Step 1: Write the balanced chemical equation and calculate the moles of reactants}$\n\n$\text{The reaction between } \text{I}_2 \text{ and } \text{Cl}_2 \text{ can be represented as:}$\n\n$\text{I}_2 + \text{Cl}_2 \rightarrow 2\text{ICl}$\n\n$\text{I}_2 + 3\text{Cl}_2 \rightarrow 2\text{ICl}_3$\n\n$\text{These can be combined into a single equation:}$\n\n$\text{I}_2 + 2\text{Cl}_2 \rightarrow \text{ICl} + \text{ICl}_3$\n\n$\text{Molar mass of } \text{I}_2 = 2 \times 127 = 254 \text{ g/mol}$\n\n$\text{Molar mass of } \text{Cl}_2 = 2 \times 35.5 = 71 \text{ g/mol}$\n\n$\text{Moles of } \text{I}_2 = \frac{25.4 \text{ g}}{254 \text{ g/mol}} = 0.1 \text{ mol}$\n\n$\text{Moles of } \text{Cl}_2 = \frac{14.2 \text{ g}}{71 \text{ g/mol}} = 0.2 \text{ mol}$\n\n$\text{Step 2: Determine the relationship between the reactants and products}$\n\n$\text{From the balanced equation, the ratio is:}$\n\n$\text{I}_2 : \text{Cl}_2 : \text{ICl} : \text{ICl}_3 = 1 : 2 : 1 : 1$\n\n$\text{Since we have 0.1 mol of } \text{I}_2 \text{ and 0.2 mol of } \text{Cl}_2\text{, the ratio of reactants matches the stoichiometric ratio exactly.}$\n\n$\text{Step 3: Calculate the moles of products formed}$\n\n$\text{According to the balanced equation, from 0.1 mol of } \text{I}_2 \text{ and 0.2 mol of } \text{Cl}_2\text{, we will form:}$\n\n$\text{Moles of ICl formed} = 0.1 \text{ mol}$\n\n$\text{Moles of } \text{ICl}_3 \text{ formed} = 0.1 \text{ mol}$\n\n$\text{Therefore, the moles of ICl and } \text{ICl}_3 \text{ formed are 0.1 and 0.1 respectively.}$

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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
}