What volume of `H_(2)` at NTP is required to convert 2.8 g of `N_(2)` into `NH_(3)`?

Answer

To solve the problem of determining the volume of hydrogen gas (H₂) required to convert 2.8 g of nitrogen gas (N₂) into ammonia (NH₃) at normal temperature and pressure (NTP), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between nitrogen and hydrogen to form ammonia can be represented by the balanced equation:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Calculate the number of moles of nitrogen (N₂)
The molar mass of nitrogen (N₂) is 28 g/mol. To find the number of moles of nitrogen in 2.8 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Number of moles of } N_2 = \frac{2.8 \, \text{g}}{28 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 3: Use the stoichiometry of the reaction
From the balanced equation, we see that 1 mole of N₂ reacts with 3 moles of H₂. Therefore, the number of moles of hydrogen required can be calculated as follows:
\[ \text{Moles of } H_2 = 3 \times \text{Moles of } N_2 \]
\[ \text{Moles of } H_2 = 3 \times 0.1 \, \text{mol} = 0.3 \, \text{mol} \]

### Step 4: Calculate the volume of hydrogen gas at NTP
At normal temperature and pressure (NTP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of hydrogen required can be calculated using:
\[ \text{Volume of } H_2 = \text{Moles of } H_2 \times 22.4 \, \text{L/mol} \]
\[ \text{Volume of } H_2 = 0.3 \, \text{mol} \times 22.4 \, \text{L/mol} = 6.72 \, \text{L} \]

### Final Answer
The volume of hydrogen gas required to convert 2.8 g of nitrogen into ammonia at NTP is **6.72 liters**.

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