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Current Question (ID: 8310)

Question:

$\text{When 4 g of iron is burnt to ferric oxide at a constant pressure, 29.28 kJ of heat is evolved.}$ $\text{The enthalpy of formation of ferric oxide will be-}$ $\text{(At. mass of Fe = 56) ?}$

Options:

1. $-81.98 \text{ kJ}$
2. $-819.8 \text{ kJ}$
3. $-40.99 \text{ kJ}$
4. $+819.8 \text{ kJ}$

Solution:

$\text{Hint: Mole concept}$ $\text{Step 1:}$ $\text{The balanced chemical equation is as follows:}$ $\text{2Fe} + \frac{3}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$ $\text{Number of moles of Fe} = \frac{4\text{ g}}{56\text{ g/mol}} = 0.0714 \text{ mol}$ $\text{2 moles of Fe gives one mole of Fe}_2\text{O}_3$ $\text{0.0714 moles of Fe will give } \frac{0.0714}{2} = 0.0357 \text{ moles of Fe}_2\text{O}_3$ $\text{Step 2:}$ $\text{When 0.0357 moles of Fe}_2\text{O}_3 \text{ are formed, 29.28 kJ of heat is evolved}$ $\text{When 1 mole of Fe}_2\text{O}_3 \text{ is formed, } 29.28 \text{ kJ} \times \frac{1 \text{ mol}}{0.0357 \text{ mol}} = 819.8 \text{ kJ/mol of heat is evolved}$ $\text{A negative sign will be introduced because heat is evolved (exothermic reaction).}$ $\text{Therefore, the enthalpy of formation of ferric oxide is } -819.8 \text{ kJ/mol}$ $\text{Hence, the answer is option 2: } -819.8 \text{ kJ}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}