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Current Question (ID: 7417)

Question:

$\text{1 g of magnesium is burnt with 0.56g of oxygen in a closed vessel. The left-out reactant and its quantity are}$ $(\text{At. weight of Mg} = 24, \text{O}=16)$

Options:

1. $\text{Mg, 0.16g}$
2. $\text{O}_2\text{, 0.16g}$
3. $\text{Mg, 0.44g}$
4. $\text{O}_2\text{, 0.28g}$

Solution:

$\text{Hint: Limiting reagent concept}$ $\text{Step 1:}$ $\text{Find the limiting reagent}$ $(1) \text{ The balanced chemical equation is}$ $\text{Mg} + \frac{1}{2} \text{O}_2 \rightarrow \text{MgO}$ $24\text{g} \quad 16\text{g} \quad 40\text{g}$ $\text{From the above equation, it is clear that 24g of Mg reacts with 16g of O}_2.$ $\text{Thus, 1.0 g of Mg reacts with 16/24g of O}_2 = 0.67\text{g of O}_2$ $\text{But only 0.56g of O}_2 \text{ is available which is less than 0.67g. Thus, O}_2 \text{ is the limiting reagent. Further, 16g of O}_2 \text{ reacts with 24g of Mg.}$ $\text{Step 2:}$ $\text{Calculate the amount of reactant left in excess}$ $0.56\text{g of O}_2 \text{ will react with Mg} = 24 \times 0.56/16 = 0.84\text{g}$ $\text{Amount of Mg left unreacted} = (1.0-0.84)\text{g Mg} = 0.16\text{g of Mg}$

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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
}