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Current Question (ID: 7400)

Question:

$\text{If 10 volumes of } \text{H}_2 \text{ gas react with 5 volumes of } \text{O}_2 \text{ gas, the volumes of water vapor produced would be:}$

Options:

1. $9$
2. $8$
3. $10$
4. $11$

Solution:

$\text{HINT: See the stoichiometry from the balanced equation.}$\n\n$\text{Step 1: Write the balanced chemical equation for the reaction of dihydrogen with dioxygen}$\n\n$2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O}\text{(g)}$\n\n$\text{Step 2: Analyze the stoichiometry from the balanced equation}$\n\n$\text{According to the balanced equation:}$\n\n$2 \text{ volumes of } \text{H}_2 \text{ react with } 1 \text{ volume of } \text{O}_2 \text{ to produce } 2 \text{ volumes of water vapor}$\n\n$\text{Step 3: Calculate the volumes of water vapor produced}$\n\n$\text{Given:}$\n\n$10 \text{ volumes of } \text{H}_2 \text{ gas and } 5 \text{ volumes of } \text{O}_2 \text{ gas}$\n\n$\text{From the stoichiometry:}$\n\n$\text{For } 2 \text{ volumes of } \text{H}_2\text{, } 1 \text{ volume of } \text{O}_2 \text{ is required}$\n\n$\text{For } 10 \text{ volumes of } \text{H}_2\text{, } \frac{10}{2} = 5 \text{ volumes of } \text{O}_2 \text{ are required}$\n\n$\text{Since we have exactly 5 volumes of } \text{O}_2\text{, all reactants will be consumed completely.}$\n\n$\text{Now, from the balanced equation:}$\n\n$2 \text{ volumes of } \text{H}_2 \text{ produce } 2 \text{ volumes of } \text{H}_2\text{O}$\n\n$10 \text{ volumes of } \text{H}_2 \text{ will produce } \frac{10}{2} \times 2 = 10 \text{ volumes of } \text{H}_2\text{O}$\n\n$\text{Therefore, the volumes of water vapor produced would be 10.}$

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
}