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Current Question (ID: 7432)

Question:

The mass of $\text{CaCO}_3$ required to react completely with 25 mL of 0.75 M $\text{HCl}$ according to the given reaction would be:\n\n$\text{CaCO}_{3(s)} + \text{HCl}_{(aq)} \rightarrow \text{CaCl}_{2(aq)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$

Options:

1. 0.36 g
2. 0.09 g
3. 0.96 g
4. 0.66 g

Solution:

\text{HINT: Find out the mass of } \text{CaCO}_3 \text{ using molarity and stoichiometry.}\n\n\text{STEP 1: Find amount of HCl present in 25 mL of solution using molarity.}\n\n0.75 \text{ M of HCl} \equiv 0.75 \text{ mol of HCl are present in 1 L of water}\n\n\equiv 27.375 \text{ g of HCl is present in 1 L of water}\n\n\text{Thus, 1000 mL of solution contains 27.375 g of HCl.}\n\n\therefore \text{ Amount of HCl present in 25 mL of solution}\n= \frac{27.375\text{g}}{1000} \text{ mL} \times 25 \text{ mL} = 0.6844 \text{ g}\n\n\text{STEP 2: Find out the amount of } \text{CaCO}_3 \text{ using stoichiometry.}\n\n\text{From the given chemical equation,}\n\text{CaCO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{CaCl}_{2(aq)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}\n\n71 \text{ g HCl react with 100 g } \text{CaCO}_3\text{.}\n\n\therefore\text{Amount of } \text{CaCO}_3 \text{ that will react with 0.6844 g} = \frac{100}{71} \times 0.6844 \text{ g}\n= 0.96 \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
}