Last week we drew Lewis dot structures of azide (N31-), a crucial component in automobile collision safety airbags. During an impact, a sensor ignites a detonator cap that sets off the following rapid reaction:
2NaN3(s) → 2Na(s) + 3N2(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? (I have done already)
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?
(unanswered)
You've done part A. Amount in part A/0.83 = amount you need at 83% efficiency.
To solve Part B of the question, we need to consider the actual yield of nitrogen gas and calculate the mass of sodium azide required to account for this yield.
The given reaction equation is:
2NaN3(s) → 2Na(s) + 3N2(g)
From the balanced equation, we can see that 2 moles of sodium azide (NaN3) produce 3 moles of nitrogen gas (N2). Therefore, the stoichiometric ratio is 2:3.
To find the mass of sodium azide required, we need the molar mass of sodium azide (NaN3). Sodium atomic mass (Na) is 22.99 g/mol, and nitrogen atomic mass (N) is 14.01 g/mol. Since there are three nitrogen atoms in sodium azide, the molar mass of sodium azide is:
Molar mass of NaN3 = (3 x 14.01 g/mol) + (1 x 22.99 g/mol)
Molar mass of NaN3 = 63.03 g/mol
Now, let's solve the problem step by step:
Step 1: Determine the moles of nitrogen gas needed.
Using the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm (1.06 atm)
V = volume in liters (65.0 L)
n = moles of gas (to be calculated)
R = gas constant (0.0821 L•atm/mol•K)
T = temperature in Kelvin (20.0 ºC = 20.0 + 273.15 K)
Rearranging the equation and solving for n:
n = (PV) / (RT)
n = (1.06 atm * 65.0 L) / (0.0821 L•atm/mol•K * 293.15 K)
n ≈ 2.78 moles of nitrogen gas
Step 2: Calculate the theoretical yield of sodium azide.
Since the stoichiometric ratio is 2:3, the moles of sodium azide required will be (2/3) times the moles of nitrogen gas.
moles of sodium azide = (2/3) * 2.78 moles
moles of sodium azide ≈ 1.85 moles
Step 3: Convert moles to grams.
To convert moles of sodium azide to grams, we need to use the molar mass of NaN3.
mass of sodium azide = moles of sodium azide * molar mass of NaN3
mass of sodium azide ≈ 1.85 moles * 63.03 g/mol
mass of sodium azide ≈ 116.66 g
Therefore, approximately 116.66 grams of sodium azide should be loaded into the device to ensure full inflation of the airbag with an expected 83% yield of nitrogen gas.