Alright, let's break down the problem step-by-step to calculate the mass of sodium azide (NaN₃) required to liberate 50.0 dm³ of nitrogen gas (N₂) at room temperature and pressure.
### Step 1: Understand the Reaction
The given decomposition reaction is:
[tex]\[ 2 \text{NaN}_3 \rightarrow 3 \text{N}_2 + 2 \text{Na} \][/tex]
### Step 2: Convert Volume of N₂ to Moles
We are given:
- Volume of nitrogen gas (N₂): 50.0 dm³
- Molar volume of gas at room temperature and pressure: 24.4 dm³/mol
Using the molar volume, we calculate the moles of nitrogen gas (N₂) liberated:
[tex]\[ \text{Moles of N}_2 = \frac{\text{Volume of N}_2}{\text{Molar volume}} \][/tex]
[tex]\[ \text{Moles of N}_2 = \frac{50.0 \, \text{dm}^3}{24.4 \, \text{dm}^3/\text{mol}} = 2.049 \, \text{moles} \][/tex]
### Step 3: Relate Moles of N₂ to Moles of NaN₃
From the balanced chemical equation, 2 moles of NaN₃ produce 3 moles of N₂. Therefore, we can set up a ratio to find the moles of NaN₃ needed:
[tex]\[ \text{Moles of NaN}_3 = \left( \frac{2}{3} \right) \times \text{Moles of N}_2 \][/tex]
[tex]\[ \text{Moles of NaN}_3 = \left( \frac{2}{3} \right) \times 2.049 = 1.366 \, \text{moles} \][/tex]
### Step 4: Calculate the Molar Mass of Sodium Azide (NaN₃)
Using the relative atomic masses:
- Na: 23 g/mol
- N: 14 g/mol
The molar mass of NaN₃ is:
[tex]\[ \text{Molar mass of NaN}_3 = 2 \times (\text{Molar mass of Na}) + 3 \times (\text{Molar mass of N}) \][/tex]
[tex]\[ \text{Molar mass of NaN}_3 = 2 \times 23 + 3 \times 14 \][/tex]
[tex]\[ \text{Molar mass of NaN}_3 = 46 + 42 = 88 \, \text{g/mol} \][/tex]
### Step 5: Calculate the Mass of NaN₃ Required
Now, we use the moles of NaN₃ and its molar mass to find the mass:
[tex]\[ \text{Mass of NaN}_3 = \text{Moles of NaN}_3 \times \text{Molar mass of NaN}_3 \][/tex]
[tex]\[ \text{Mass of NaN}_3 = 1.366 \, \text{moles} \times 88 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of NaN}_3 = 120.219 \, \text{g} \][/tex]
### Conclusion
The mass of sodium azide required to liberate 50.0 dm³ of nitrogen gas at room temperature is approximately 120.22 g.
