Import Question JSON

Current Question (ID: 20793)

Question:

$\text{Which of the following statements is true regarding the reaction given below?}$ $2\text{Al}_{(s)} + 6\text{HCl}_{(aq)} \rightarrow 2\text{Al}^{3+}_{(aq)} + 6\text{Cl}^{-}_{(aq)} + 3\text{H}_2_{(g)}$

Options:

1. $6\text{L HCl}_{(aq)} \text{ is consumed for every } 3\text{L of H}_2_{(g)} \text{ produced.}$
2. $33.6\text{ L H}_2_{(g)} \text{ is produced regardless of temperature and pressure for every mole of Al that reacts.}$
3. $67.2\text{ L H}_2_{(g)} \text{ at STP is produced for every mole of Al that reacts.}$
4. $11.2\text{ L H}_2_{(g)} \text{ at STP is produced for every mole of HCl}_{(aq)} \text{ consumed.}$

Solution:

$\text{Hint: 1 mole of an ideal gas occupies 22.4 L at STP.}$ $\text{Explanation:}$ $\text{The given equation is:}$ $2\text{Al}_{(s)} + 6\text{HCl}_{(aq)} \rightarrow 2\text{Al}^{3+}_{(aq)} + 6\text{Cl}^{-}_{(aq)} + 3\text{H}_2_{(g)}$ $\text{This equation indicates that 2 moles of solid aluminum react with 6 moles of hydrochloric acid in aqueous solution to produce 2 moles of aluminum ions in aqueous solution, 6 moles of chloride ions in aqueous solution, and 3 moles of hydrogen gas.}$ $(1) \text{ According to the stoichiometry of the balanced reaction, 6 moles of HCl react to produce 3 moles of } H_2. \text{ We can't directly write volume in place of number of moles.}$ $(2) \text{ 1 mole of } Al \text{ produces 1.5 moles of } H_2 \text{ or } 1.5 \times 22.4\text{ L } = 33.6\text{ L of } H_2 \text{ at STP. It can't have the same volume at any temperature or pressure condition.}$ $(3) \text{ 1 mole of } HCl \text{ produces 0.5 moles of } H_2 \text{ or } 0.5 \times 22.4\text{ L } = 11.2\text{ L of } H_2 \text{ at STP.}$ $\text{Thus, option 4 is the correct answer.}$

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
}