Import Question JSON

Current Question (ID: 7411)

Question:
$\text{Amount of HCl that would react with } 5.0 \text{ g of manganese dioxide, as per the given reaction will be:}$ $4\text{HCl}_{(aq)} + \text{MnO}_{2(s)} \rightarrow 2\text{H}_2\text{O}_{(l)} + \text{MnCl}_{2(aq)} + \text{Cl}_{2(g)}$
Options:
  • 1. $4.8 \text{ g}$
  • 2. $6.4 \text{ g}$
  • 3. $2.8 \text{ g}$
  • 4. $8.4 \text{ g}$
Solution:
$\text{HINT: Equation based problem. Solve it using balanced equation.}$ $\text{From balanced equation, we can see that}$ $1 \text{ mol } [55 + 2 \times 16 = 87 \text{ g}] \text{ MnO}_2 \text{ reacts completely with } 4 \text{ mol } [4 \times 36.5 = 146 \text{ g}] \text{ of HCl.}$ $5.0 \text{ g of MnO}_2 \text{ will react with}$ $= \frac{146}{87} \times 5.0 \text{ g of HCl}$ $= 8.4 \text{ g of HCl}$ $\text{Hence, } 8.4 \text{ g of HCl will react completely with } 5.0 \text{ g of manganese dioxide.}$

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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
}