17 g of `NH_(3)` gas will occupy a volume of _____________ `cm^(3)` at NTP.

Answer

To solve the problem of how much volume 17 g of NH₃ gas will occupy at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Determine the molar mass of NH₃
The molar mass of ammonia (NH₃) can be calculated as follows:
- Nitrogen (N) has a molar mass of approximately 14 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol, and there are three hydrogen atoms in NH₃.

Thus, the molar mass of NH₃ is:
\[ \text{Molar mass of NH}_3 = 14 \, \text{g/mol} + (3 \times 1 \, \text{g/mol}) = 14 \, \text{g/mol} + 3 \, \text{g/mol} = 17 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of NH₃ in 17 g
Using the molar mass, we can find the number of moles of NH₃ in 17 g:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{17 \, \text{g}}{17 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Use the volume occupied by 1 mole of gas at NTP
At NTP, 1 mole of any ideal gas occupies a volume of 22,400 mL (or cm³). Since we have calculated that there is 1 mole of NH₃, we can directly use this information.

### Step 4: Calculate the volume occupied by 1 mole of NH₃
Since we have 1 mole of NH₃, the volume it occupies at NTP is:
\[ \text{Volume} = 1 \, \text{mol} \times 22,400 \, \text{mL/mol} = 22,400 \, \text{mL} \]

### Conclusion
Thus, 17 g of NH₃ gas will occupy a volume of **22,400 cm³** at NTP.

---