Last week we drew Lewis dot structures of azide (N_3^{1-}N
3
1−
), a crucial component in automobile collision safety airbags. During an impact, a sensor ignites a detonator cap that sets off the following rapid reaction:
2NaN_{3(s)} \rightarrow 2Na_{(s)} + 3N_{2(g)}2NaN
3(s)
→2Na
(s)
+3N
2(g)
Part A
What mass of sodium azide must be completely reacted to inflate an air bag to 65.0 liters at 20.0 ºC and 1.06 atm?
Part B
What mass of sodium azide should be loaded in the device to ensure full inflation of the same air bag if the actual reaction can be expected to proceed to give only 83% yield of nitrogen gas?
To learn more about airbags and the chemistry behind how they work, look here:
Bell, W.L. "Chemistry of Air Bags," (1990) J. Chem. Ed. 67 (1), p. 61.
Part A:
Well, I hope this doesn't turn into a "deflated" situation. To solve this problem, we need to use the ideal gas law.
First, let's convert the volume to moles using the ideal gas law equation: PV = nRT.
Since we have the pressure, volume, and temperature, we can rearrange the equation to solve for the number of moles.
Let's convert the temperature to Kelvin by adding 273 to 20 degrees Celsius. So, the temperature is 293 K.
Now we can plug in the values into the equation:
(1.06 atm)(65.0 L) = n(0.0821 L·atm/(K·mol))(293 K)
Solving for n, we get:
n = (1.06 atm × 65.0 L) / (0.0821 L·atm/(K·mol) × 293 K)
Now we can calculate the number of moles of nitrogen gas produced using the stoichiometry of the reaction.
The balanced equation tells us that 2 moles of sodium azide produce 3 moles of nitrogen gas. So, we need to convert moles of nitrogen gas to moles of sodium azide.
Since we know the molar mass of sodium azide is 65 g/mol, we can calculate the mass of sodium azide needed.
So, the mass of sodium azide required is:
mass = n(molar mass)
Now, I'll leave the calculations to you, and remember to always wear your "chemistry goggles" for safety!
To solve these problems, we need to use the ideal gas law and stoichiometry. Let's begin step-by-step:
Part A:
1. We'll start by converting the given volume and temperature into standard conditions (STP) to make calculations easier. Standard conditions are defined as 0 ºC (273.15 K) and 1 atm.
We can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
Step 1: Convert the temperature from Celsius to Kelvin.
20.0 ºC + 273.15 = 293.15 K
Step 2: Calculate the number of moles (n) using the ideal gas law equation.
n = PV / RT
n = (1.06 atm) * (65.0 L) / (0.0821 L·atm/mol·K * 293.15 K)
2. Now we need to determine the stoichiometric relationship between sodium azide (NaN3) and nitrogen gas (N2) using the balanced equation.
According to the balanced equation: 2 NaN3 → 2 Na + 3 N2.
This means that for every 2 moles of NaN3 reacted, we will produce 3 moles of N2.
3. Convert moles of NaN3 to grams using the molar mass of NaN3.
The molar mass of NaN3 = 65.01 g/mol.
mass of NaN3 = n * molar mass of NaN3
Substitute the value of n calculated in step 1 and calculate the mass of NaN3 required.
Part B:
1. Calculate the theoretical yield of nitrogen gas (N2) using the stoichiometry and the molar ratio of nitrogen gas to sodium azide from the balanced equation.
In this case, we have 83% yield, meaning 83% of the expected theoretical yield will be produced. So we need to divide the desired mass by 0.83 to account for the yield.
theoretical yield = (mass of N2) / 0.83
2. Calculate the mass of sodium azide (NaN3) needed using the stoichiometry and the molar ratio of sodium azide to nitrogen gas from the balanced equation.
Using the theoretical yield of nitrogen gas calculated in step 1, we can calculate the mass of sodium azide required.
mass of NaN3 = (theoretical yield of N2) * (molar mass of NaN3)
Now we can plug in the given values and solve the equations to find the answers for both parts.
To solve Part A of the problem, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, we need to convert the volume from liters to moles using the ideal gas law equation. Rearranging the equation, we have:
n = PV / RT
Given:
V = 65.0 L
P = 1.06 atm
T = 20.0 ºC = 293.15 K
Plugging in these values, we can calculate the number of moles of nitrogen gas produced:
n = (1.06 atm * 65.0 L) / (0.0821 L·atm/mol·K * 293.15 K)
n = 3.59 moles
Since the balanced chemical equation tells us that for every 2 moles of NaN3, we get 3 moles of N2 gas, we can calculate the number of moles of NaN3 needed:
2 moles of NaN3 --------------- 3 moles of N2
x moles of NaN3 --------------- 3.59 moles of N2
Solving for x, we get:
x = (2 moles of NaN3 * 3.59 moles of N2) / 3 moles of N2
x = 2.39 moles of NaN3
To find the mass of sodium azide, we can use its molar mass, which is the sum of the atomic masses of its constituent elements:
Na: 22.99 g/mol
N: 14.01 g/mol
Molar mass of NaN3 = (3 * 14.01 g/mol) + 22.99 g/mol = 65.02 g/mol
Now, we can calculate the mass of sodium azide needed:
Mass = moles * molar mass
Mass = 2.39 moles * 65.02 g/mol
Mass ≈ 155.57 g
Therefore, approximately 155.57 grams of sodium azide must be completely reacted to inflate an airbag to 65.0 liters at 20.0 ºC and 1.06 atm.
Moving on to Part B of the problem, we need to consider the yield of the reaction. The problem states that the reaction can be expected to have an 83% yield of nitrogen gas.
To calculate the mass of sodium azide needed to ensure full inflation, we need to adjust the number of moles of NaN3:
Total moles of N2 needed = (3.59 moles of N2) / 0.83
Total moles of N2 needed ≈ 4.33 moles
Using the same stoichiometry as before, we can calculate the number of moles of sodium azide needed:
x = (2 moles of NaN3 * 4.33 moles of N2) / 3 moles of N2
x ≈ 2.89 moles of NaN3
As we calculated previously, the molar mass of NaN3 is 65.02 g/mol. Using this molar mass, we can calculate the mass of sodium azide needed:
Mass = moles * molar mass
Mass = 2.89 moles * 65.02 g/mol
Mass ≈ 187.86 g
Therefore, approximately 187.86 grams of sodium azide should be loaded in the device to ensure full inflation of the airbag considering an 83% yield of nitrogen gas.