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Question:
\text{Balance the following reaction and find the values of a, b and g, respectively:} \\[0.5em] a\text{KMnO}_4 + b\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow d\text{MnSO}_4 + e\text{K}_2\text{SO}_4 + f\text{O}_2 + g\text{H}_2\text{O}
Options:
  • 1. $ 2, 4, 8 $
  • 2. $ 2, 5, 8 $
  • 3. $ 10, 4, 16 $
  • 4. $ 8, 5, 2 $
Solution:
\text{Hint: Stoichiometric coefficient of KMnO}_4 \text{ is 2.} \\[1em] \text{The given reaction is as follows:} \\[0.5em] a\text{KMnO}_4 + b\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow d\text{MnSO}_4 + e\text{K}_2\text{SO}_4 + f\text{O}_2 + g\text{H}_2\text{O} \\[1em] \text{First, balance the potassium atom by multiplying KMnO}_4 \text{ by 2.} \\[0.5em] 2\text{KMnO}_4 + b\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow d\text{MnSO}_4 + 1\text{K}_2\text{SO}_4 + f\text{O}_2 + g\text{H}_2\text{O} \\[1em] \text{Now, balance the Mn atoms. Since there are 2 Mn on the left side (from 2 KMnO}_4\text{),} \\ \text{we need 2 MnSO}_4 \text{ on the right side. Thus, d=2.} \\[0.5em] 2\text{KMnO}_4 + b\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow 2\text{MnSO}_4 + 1\text{K}_2\text{SO}_4 + f\text{O}_2 + g\text{H}_2\text{O} \\[1em] \text{Now, balance the sulfur atoms. On the right side, there are 2 S from 2 MnSO}_4 \\ \text{and 1 S from K}_2\text{SO}_4 \text{ (total 3 S). This matches the 3 H}_2\text{SO}_4 \text{ on the left side,} \\ \text{so the coefficient for H}_2\text{SO}_4 \text{ is correct as 3.} \\[1em] \text{The balanced equation is:} \\[0.5em] 2\text{KMnO}_4 + 5\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow 2\text{MnSO}_4 + \text{K}_2\text{SO}_4 + 5\text{O}_2 + 8\text{H}_2\text{O} \\[1em] \text{Therefore: } a=2, \; b=5, \; g=8

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{
  "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
}