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Current Question (ID: 20814)

Question:
$\text{Consider a mixture of } \text{CH}_4 \text{ and } \text{C}_2\text{H}_4 \text{ having a volume of } 16.8 \text{ L at } 273 \text{ K and } 1 \text{ atm. It undergoes combustion to form } \text{CO}_2 \text{ with a total volume of } 28 \text{ L at the same temperature and pressure.}$ $\text{If the enthalpy of combustion of } \text{CH}_4 \text{ is } -900 \text{ kJ/mol and enthalpy of combustion of } \text{C}_2\text{H}_4 \text{ is } -1400 \text{ kJ/mol, then the magnitude of heat released on combustion of the given mixture in kJ is-}$
Options:
  • 1. $875 \text{ kJ}$
  • 2. $925 \text{ kJ}$
  • 3. $945 \text{ kJ}$
  • 4. $900 \text{ kJ}$
Solution:
$\text{Hint: At STP, volume of 1 mole of any gas is } 22.4\text{L.}$ $\text{Step 1: The total volume of mixture of } \text{CH}_4 \text{ and } \text{C}_2\text{H}_4 \text{ is } 16.8\text{L. Let the volume of } \text{CH}_4 \text{ is } \text{''}x\text{'' L}$ $\text{The combustion reaction of } \text{CH}_4 \text{ and } \text{C}_2\text{H}_4 \text{ are:}$ $\text{(i) } \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ $x$ $\text{(ii) } \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}$ $2(16.8-x)$ $\text{Step 2: The total volume of } \text{CO}_2 \text{ after combustion = } 28\text{L}$ $x + 2(16.8-x) = 28\text{L}$ $\text{Thus, } x = 5.6\text{L}$ $\text{Step 3: Number of moles of } \text{CH}_4 = \frac{5.6}{22.4} = \frac{1}{4}$ $\text{Similarly, number of moles of } \text{C}_2\text{H}_4 = \frac{16.8-5.6}{22.4} = \frac{1}{2}$ $\text{Hence,}$ $\text{Heat released } = \frac{1}{4} \times 900 + \frac{1}{2} \times 1400 = 225 + 700$ $= 925 \text{ kJ}$ $\text{So, option 2 is the correct answer.}$

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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
}