Import Question JSON

Current Question (ID: 8308)

Question:

$\text{The enthalpy of combustion of carbon to CO}_2 \text{ is } -393.5 \text{ kJ mol}^{-1}\text{. The amount of heat released upon formation of } 35.2 \text{ g of CO}_2 \text{ from carbon and dioxygen gas would be:}$

Options:

1. $-393.5 \text{ kJ mol}^{-1}$
2. $-314.8 \text{ kJ mol}^{-1}$
3. $+314.8 \text{ kJ mol}^{-1}$
4. $-320.5 \text{ kJ mol}^{-1}$

Solution:

$\text{Hint: Use the unitary method to solve this question.}$ $\text{Explanation:}$ $\text{Step 1: Find the heat released by one mole of CO}_2$ $\text{For the reaction: C}_{(s)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} \text{; } \Delta_f H = -393.5 \text{ kJ mol}^{-1}$ $\text{Step 2: Find the heat released by } 35.2 \text{ g CO}_2$ $\text{First, we need to convert the mass of CO}_2 \text{ to moles:}$ $\text{Molar mass of CO}_2 = 12 + 2 \times 16 = 44 \text{ g/mol}$ $\text{Number of moles of CO}_2 = \frac{35.2 \text{ g}}{44 \text{ g/mol}} = 0.8 \text{ mol}$ $\text{Heat released on formation of 1 mol CO}_2 = -393.5 \text{ kJ}$ $\text{Heat released on formation of } 0.8 \text{ mol CO}_2 = -393.5 \text{ kJ/mol} \times 0.8 \text{ mol} = -314.8 \text{ kJ}$ $\text{Therefore, the amount of heat released upon formation of } 35.2 \text{ g of CO}_2 \text{ would be } -314.8 \text{ kJ.}$ $\text{The negative sign indicates that heat is released (exothermic process).}$

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
}