Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide (NaN3) to decompose explosively according to the following reaction.
2 NaN3(s) 2 Na(s) + 3 N2(g)
What mass of NaN3(s) must be reacted in order to inflate an air bag to 73.1 L at STP?
Is STP T=273K and P=1atm?
Use PV = nRT, substitute the numbers and solve for n = number of moles N2.
Using the coefficients in the balanced equation, convert moles N2 to moles NaN3. Convert moles NaN3 to grams. g = moles x molar mass.
Post your work if you get stuck.
To find the mass of NaN3(s) needed to inflate an airbag to 73.1 L at STP, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP is 1 atm)
V = volume (73.1 L)
n = moles of NaN3
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (STP is 273.15 K)
We need to solve for the moles of NaN3 first and then convert it to grams using the molar mass.
Step 1: Calculate the moles of N2 gas produced
From the balanced equation, we can see that 2 moles of NaN3 produce 3 moles of N2 gas. Therefore, the moles of NaN3 will be the same as the moles of N2 gas produced when the airbag inflates.
n = (2 moles of NaN3) * (3 moles of N2 gas / 2 moles of NaN3) = 3 moles of N2 gas
Step 2: Calculate the volume of N2 gas at STP
We can use the ideal gas law equation to calculate the volume of N2 gas at STP:
PV = nRT
(1 atm) * (V) = (3 moles) * (0.0821 L•atm/mol•K) * (273.15 K)
V = (3 moles) * (0.0821 L•atm/mol•K) * (273.15 K) / (1 atm)
V = 67.74 L
Step 3: Convert the volume of N2 gas to moles of NaN3
Since the volume of N2 gas is equal to the volume of the airbag, we can use the following equation:
(67.74 L) * (x moles of NaN3) / (73.1 L) = 3 moles of N2 gas
x = (3 moles) * (73.1 L) / (67.74 L) = 3.25 moles of NaN3
Step 4: Calculate the mass of NaN3
The molar mass of NaN3 is the sum of the atomic masses of sodium (Na) and nitrogen (N), multiplied by the number of each element in the formula:
Molar mass of NaN3 = (22.99 g/mol + 14.01 g/mol * 3) = 65.01 g/mol
Mass = (3.25 moles) * (65.01 g/mol) = 211.14 g
Therefore, approximately 211.14 grams of NaN3 must be reacted to inflate an airbag to 73.1 L at STP.
To find the mass of NaN3 needed to inflate an airbag to a specific volume, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP pressure is 1 atm)
V = volume (73.1 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
First, we need to calculate the number of moles of N2 gas produced by the decomposition reaction of NaN3. According to the balanced equation, 2 moles of NaN3 produce 3 moles of N2.
Given that the volume of the airbag at STP is 73.1 L, we convert this to moles using the ideal gas law:
n = PV / RT
n = (1 atm) * (73.1 L) / (0.0821 L·atm/mol·K * 273.15 K)
n = 3.055 moles of N2 gas
Since the stoichiometry of the reaction is 2 moles of NaN3 to 3 moles of N2, we can calculate the moles of NaN3 required:
3 moles of N2 correspond to (2/3) * 3.055 moles of NaN3
n(NaN3) = 2.037 moles of NaN3
Now we need to convert moles of NaN3 to grams. The molar mass of NaN3 is:
Molar mass of NaN3 = (1 * 22.99 g/mol) + (3 * 14.00 g/mol) = 65.99 g/mol
Mass of NaN3 = n(NaN3) * Molar mass of NaN3
Mass of NaN3 = 2.037 moles * 65.99 g/mol
Therefore, the mass of NaN3 that must be reacted in order to inflate the airbag to 73.1 L at STP is approximately 134.62 grams.