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Sagot :
Let's determine the mass of the remaining excess reagent by following a step-by-step method.
### Step 1: Determine the molar masses
- Molar mass of \( NH_3 \) (ammonia):
- \( N = 14.01 \, \text{g/mol} \)
- \( H = 3 \times 1.01 \, \text{g/mol} \)
- Total molar mass of \( NH_3 = 14.01 + 3 \times 1.01 = 17.03 \, \text{g/mol} \)
- Molar mass of \( O_2 \) (oxygen gas):
- \( O = 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \)
### Step 2: Calculate the number of moles of \( NH_3 \) and \( O_2 \)
- Moles of \( NH_3 \):
- Given mass of \( NH_3 \) is \( 40.0 \, \text{g} \)
- \( \text{Moles of } NH_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{40.0 \, \text{g}}{17.03 \, \text{g/mol}} \approx 2.348 \, \text{moles} \)
- Moles of \( O_2 \):
- Given mass of \( O_2 \) is \( 50.0 \, \text{g} \)
- \( \text{Moles of } O_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{50.0 \, \text{g}}{32.00 \, \text{g/mol}} \approx 1.5625 \, \text{moles} \)
### Step 3: Determine the limiting reagent
The stoichiometric coefficients from the balanced equation \( 4 NH_3 + 5 O_2 \rightarrow 4 NO + 6 H_2O \) are:
- 4 moles of \( NH_3 \)
- 5 moles of \( O_2 \)
Calculate the stoichiometric ratios:
- \( \text{Ratio for } NH_3 = \frac{\text{Moles of } NH_3}{4} = \frac{2.348}{4} \approx 0.587 \)
- \( \text{Ratio for } O_2 = \frac{\text{Moles of } O_2}{5} = \frac{1.5625}{5} \approx 0.3125 \)
Since \( 0.3125 \) (ratio of \( O_2 \)) is smaller than \( 0.587 \) (ratio of \( NH_3 \)), \( O_2 \) is the limiting reagent.
### Step 4: Calculate the mass of the excess reagent (remaining \( NH_3 \))
- Moles of \( NH_3 \) that react with the limiting reagent (\( O_2 \)):
- According to the stoichiometry, \( 5 \) moles of \( O_2 \) react with \( 4 \) moles of \( NH_3 \)
- \( \text{Moles of } NH_3 \text{ reacting with 1.5625 moles of } O_2 = \frac{4}{5} \times 1.5625 = 1.25 \, \text{moles} \)
- Moles of \( NH_3 \) left:
- \( \text{Total moles of } NH_3 - \text{Moles of } NH_3 \text{ that reacted} = 2.348 - 1.25 \approx 1.098 \, \text{moles} \)
- Mass of remaining \( NH_3 \):
- \( \text{Mass of remaining } NH_3 = \text{Moles of remaining } NH_3 \times \text{Molar mass of } NH_3 \)
- \( \text{Mass of remaining } NH_3 = 1.098 \, \text{moles} \times 17.03 \, \text{g/mol} \approx 18.7125 \, \text{g} \)
### Conclusion:
The mass of the remaining excess reagent is approximately \( 18.7 \, \text{g} \).
So, the correct answer is c. 18.7 g.
### Step 1: Determine the molar masses
- Molar mass of \( NH_3 \) (ammonia):
- \( N = 14.01 \, \text{g/mol} \)
- \( H = 3 \times 1.01 \, \text{g/mol} \)
- Total molar mass of \( NH_3 = 14.01 + 3 \times 1.01 = 17.03 \, \text{g/mol} \)
- Molar mass of \( O_2 \) (oxygen gas):
- \( O = 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \)
### Step 2: Calculate the number of moles of \( NH_3 \) and \( O_2 \)
- Moles of \( NH_3 \):
- Given mass of \( NH_3 \) is \( 40.0 \, \text{g} \)
- \( \text{Moles of } NH_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{40.0 \, \text{g}}{17.03 \, \text{g/mol}} \approx 2.348 \, \text{moles} \)
- Moles of \( O_2 \):
- Given mass of \( O_2 \) is \( 50.0 \, \text{g} \)
- \( \text{Moles of } O_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{50.0 \, \text{g}}{32.00 \, \text{g/mol}} \approx 1.5625 \, \text{moles} \)
### Step 3: Determine the limiting reagent
The stoichiometric coefficients from the balanced equation \( 4 NH_3 + 5 O_2 \rightarrow 4 NO + 6 H_2O \) are:
- 4 moles of \( NH_3 \)
- 5 moles of \( O_2 \)
Calculate the stoichiometric ratios:
- \( \text{Ratio for } NH_3 = \frac{\text{Moles of } NH_3}{4} = \frac{2.348}{4} \approx 0.587 \)
- \( \text{Ratio for } O_2 = \frac{\text{Moles of } O_2}{5} = \frac{1.5625}{5} \approx 0.3125 \)
Since \( 0.3125 \) (ratio of \( O_2 \)) is smaller than \( 0.587 \) (ratio of \( NH_3 \)), \( O_2 \) is the limiting reagent.
### Step 4: Calculate the mass of the excess reagent (remaining \( NH_3 \))
- Moles of \( NH_3 \) that react with the limiting reagent (\( O_2 \)):
- According to the stoichiometry, \( 5 \) moles of \( O_2 \) react with \( 4 \) moles of \( NH_3 \)
- \( \text{Moles of } NH_3 \text{ reacting with 1.5625 moles of } O_2 = \frac{4}{5} \times 1.5625 = 1.25 \, \text{moles} \)
- Moles of \( NH_3 \) left:
- \( \text{Total moles of } NH_3 - \text{Moles of } NH_3 \text{ that reacted} = 2.348 - 1.25 \approx 1.098 \, \text{moles} \)
- Mass of remaining \( NH_3 \):
- \( \text{Mass of remaining } NH_3 = \text{Moles of remaining } NH_3 \times \text{Molar mass of } NH_3 \)
- \( \text{Mass of remaining } NH_3 = 1.098 \, \text{moles} \times 17.03 \, \text{g/mol} \approx 18.7125 \, \text{g} \)
### Conclusion:
The mass of the remaining excess reagent is approximately \( 18.7 \, \text{g} \).
So, the correct answer is c. 18.7 g.
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