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Current Question (ID: 7416)

Question:

$\text{One mole of a mixture of CO and CO}_2 \text{ requires exactly 20g of NaOH in solution for the complete conversion of all the CO}_2 \text{ into Na}_2\text{CO}_3\text{. How much NaOH would it require for conversion into Na}_2\text{CO}_3\text{, if the mixture (one mole) is completely oxidised to CO}_2\text{?}$

Options:

1. $60\text{ g}$
2. $80\text{ g}$
3. $40\text{ g}$
4. $20\text{ g}$

Solution:

$\text{Hint: Balanced chemical equation } 2\text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2 \text{CO}_3 + \text{H}_2\text{O}$ $\text{Step 1:}$ $\text{Let moles of CO} = x$ $\text{So, moles of CO}_2 = 1-x$ $\text{Now, } 2\text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2 \text{CO}_3 + \text{H}_2\text{O}$ $2 \text{ moles NaOH will react with 1 moles of CO}_2$ $\text{So, 20g NaOH} = 0.5 \text{ moles NaOH will react with } \frac{1}{2} = 0.25 \text{ moles CO}_2$ $= 1-x \text{ moles}$ $\text{Therefore, } x = 0.75 \text{ moles is moles of CO} = 0.75 \text{ moles}$ $\text{Step 2:}$ $\text{Now, on oxidation } \rightarrow \text{CO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}_2$ $\text{CO}(0.75 \text{ moles}) \rightarrow \text{CO}_2(0.75 \text{ moles})$ $\text{Total moles of CO}_2 = (0.25 + 0.75) \text{ moles}$ $= 1 \text{ moles}$ $1 \text{ mole CO}_2 \text{ will react with 2 moles of NaOH}$ $i.e., (40 \times 2) = 80 \text{ g}$

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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
}