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Current Question (ID: 7392)

Question:
$\text{When 22.4 L of } \text{H}_2 \text{ (g) is mixed with 11.2 L of } \text{Cl}_2 \text{ (g), each at STP, the moles of } \text{HCl(g)} \text{ formed is equal to - }$
Options:
  • 1. $1 \text{ mole of } \text{HCl} \text{ (g)}$
  • 2. $2 \text{ moles of } \text{HCl} \text{ (g)}$
  • 3. $0.5 \text{ mole of } \text{HCl} \text{ (g)}$
  • 4. $1.5 \text{ mole of } \text{HCl} \text{ (g)}$
Solution:
$\text{Hint: Stoichiometry based Limiting reagent related problems.}$\n\n$\text{Step 1:}$\n\n$\text{The given problem is related to the concept of the stoichiometry of chemical equations.}$\n\n$\text{Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent}$\n\n$\text{[possessing a minimum number of moles and gets completely used up in the reaction].}$\n\n$\text{The limiting reagent gives the moles of product formed in the reaction.}$\n\n$\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl} \text{(g)}$\n\n$\text{Initial volumes: } \text{H}_2 = 22.4 \text{ L, } \text{Cl}_2 = 11.2 \text{ L}$\n\n$\text{22.4 L volume at STP is occupied by 1 mole of gas,}$\n\n$\text{So for } \text{H}_2 = 1 \text{ mole}$\n\n$\text{11.2 L volume will be occupied by,}$\n\n$\text{Cl}_2 = 1 \times 11.2/22.4 \text{ mole} = 0.5 \text{ mol}$\n\n$\text{From the equation it is clear that 1 mol } \text{H}_2 \text{ needs 1 mol } \text{Cl}_2 \text{ but the given mole of } \text{Cl}_2 \text{ is 0.5 mol. Hence, } \text{Cl}_2 \text{ is a limiting reagent.}$\n\n$\text{1 mol } \text{Cl}_2 \text{ will give 2 mol of } \text{HCl}$\n\n$\text{0.5 mol of } \text{Cl}_2 \text{ will give 1 mol of } \text{HCl}$\n\n$\text{Thus, the amount of HCl formed is 1 mole.}$

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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
}