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Current Question (ID: 20802)

Question:

$6.1$ $\text{gram of CNG on combustion with } 208 \text{ grams of } \text{O}_2(g) \text{ produces, } \text{CO}_2(g) \text{ and } \text{H}_2\text{O with a lot of heat. The amount of } \text{CO}_2 \text{ produced in grams is:}$ $[\text{Consider CNG as methane and report your answer to the nearest integer.}]$

Options:

1. $15 \text{ g}$
2. $19 \text{ g}$
3. $17 \text{ g}$
4. $21 \text{ g}$

Solution:

$\text{Hint: The reaction is } \text{CH}_4 + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}$ $\text{STEP 1:}$ $\text{The reaction between methane and } \text{O}_2 \text{ is represented as:}$ $\text{CH}_4 + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}$ $\text{The mass of CNG or methane = } 6.1 \text{g and } \text{O}_2 \text{ is } 208\text{g.}$ $\text{Now, } \text{CH}_4 + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}$ $\frac{6.1}{16} \text{ mole} \quad \frac{208}{32} \text{ mole}$ $\text{or, } 0.38125 \text{ moles of } \text{CH}_4 \text{ and } 6.5 \text{ moles of } \text{O}_2 \text{ are present. 1 mole of methane reacts with 2 moles of } \text{O}_2. \text{ Thus, } \text{CH}_4 \text{ is the limiting reagent here.}$ $\text{STEP 2:}$ $\text{According to the stoichiometry, 1 mole of } \text{CH}_4 \text{ reacts with 2 moles of } \text{O}_2 \text{ to give 1 mole of } \text{CO}_2. \text{ Thus, from } 0.3815 \text{ moles of } \text{CH}_4, 0.31825 \text{ moles of } \text{CO}_2 \text{ will be formed.}$ $\text{So, the weight of } \text{CO}_2 = 0.31825 \times 44 = 16.775\text{g or } 17\text{g}$ $\text{Thus, option 3 is the correct answer.}$

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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
}