3. Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor:

[tex]\[ 4 NH_3 + 5 O_2 \rightarrow 4 NO + 6 H_2O \][/tex]

When $40.0 \, \text{g} \, NH_3$ and $50.0 \, \text{g} \, O_2$ are allowed to react, what is the mass of the remaining excess reagent?

A. $1.25 \, \text{g}$

B. $2.35 \, \text{g}$

C. $18.7 \, \text{g}$

D. $21.3 \, \text{g}$

E. [tex]$43.9 \, \text{g}$[/tex]

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.