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Current Question (ID: 8580)

Question:

$\text{The maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen is:}$

Options:

1. $9 \text{ g}$
2. $15 \text{ g}$ (Correct)
3. $12 \text{ g}$
4. $11 \text{ g}$

Solution:

$\text{Hint: Use limiting reagent concept.}$ $\text{EXPLANATION:}$ $\text{Step 1:}$ $\text{The balanced chemical equation for the given reaction is given as:}$ $4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 4\text{NO}\text{(g)} + 6\text{H}_2\text{O}\text{(g)}$ $4 \times 17\text{g} \quad 5 \times 32\text{g} \quad 4 \times 30\text{g} \quad 6 \times 18\text{g}$ $= 68\text{g} \quad = 160\text{g} \quad = 120\text{g} \quad = 108\text{g}$ $\text{Thus, 68g of NH}_3 \text{ reacts with 160g of O}_2\text{.}$ $\text{Therefore, 10g of NH}_3 \text{ reacts with } \frac{160 \times 10}{68} \text{g of O}_2\text{, or 23.53g of O}_2\text{.}$ $\text{But the available amount of O}_2 \text{ is 20 g.}$ $\text{Therefore, O}_2 \text{ is the limiting reagent (we have considered the amount of O}_2 \text{ to calculate the weight of nitric oxide obtained in the reaction).}$ $\text{Step 2:}$ $\text{Now, 160 g of O}_2 \text{ gives 120g of NO.}$ $\text{Therefore, 20 g of O}_2 \text{ gives } \frac{120 \times 20}{160} \text{ g of NO, or 15 g of NO.}$ $\text{Hence, a maximum of 15 g of nitric oxide can be obtained.}$

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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
}