Air bags in cars are inflated by the sudden
decomposition of sodium azide (NaN3) by the
following reaction.
2 NaN3(s) −→ 3 N2(g) + 2 Na(s)
What volume of N2 gas, measured at 1.30
atm and 88.0◦C, would be produced by the
reaction of 65.0 g of NaN3?
Answer in units of L.
Well, let's calculate this together, shall we?
First, we need to determine the number of moles of NaN3. To do that, we can use its molar mass, which is 65 grams per mole. So, 65.0 g of NaN3 is equal to 1 mole of NaN3.
According to the balanced equation, for every 2 moles of NaN3, we get 3 moles of N2 gas. Therefore, if we have 1 mole of NaN3, we would get 3/2 moles of N2 gas.
Now, let's use the ideal gas law: PV = nRT. We are given the pressure (1.30 atm), temperature (88.0◦C or 361 K), and we want to find the volume (V) in liters. Also, the number of moles (n) is 3/2.
Plugging in the values:
(1.30 atm) x (V) = (3/2) x (0.0821 L•atm/mol•K) x (361 K)
Let's solve for V:
V = (3/2) x (0.0821 L•atm/mol•K) x (361 K) / (1.30 atm)
After crunching the numbers, we get:
V ≈ 13.48 L
So, approximately 13.48 liters of N2 gas would be produced. I hope this answer didn't deflate your expectations!
To calculate the volume of N2 gas produced, we'll need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.30 atm)
V = volume (unknown)
n = number of moles of gas (which we'll calculate)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (which we'll convert from °C)
First, let's calculate the number of moles of N2 gas produced:
1 mole of NaN3 produces 3 moles of N2
molar mass of NaN3 = sodium (22.99 g/mol) + azide (14.01 g/mol x 3) = 65.02 g/mol
So, 65.0 g of NaN3 is equal to (65.0 g / 65.02 g/mol) = 0.999 moles of NaN3
Since the reaction produces a 3:2 ratio of N2 to NaN3, we'll multiply the number of moles of NaN3 by (3/2) to find the number of moles of N2:
0.999 moles of NaN3 x (3/2) = 1.4985 moles of N2
Next, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 88.0°C + 273.15 = 361.15 K
Now we can solve for the volume (V) using the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (1.4985 mol x 0.0821 L·atm/(mol·K) x 361.15 K) / 1.30 atm
V ≈ 32.7 L
Therefore, approximately 32.7 liters of N2 gas would be produced by the reaction of 65.0 g of NaN3.
To find the volume of N2 gas produced by the reaction of 65.0 g of NaN3, we need to use the concept of stoichiometry and the ideal gas law.
1. Start by balancing the equation:
2 NaN3(s) → 3 N2(g) + 2 Na(s)
2. Convert the mass of NaN3 to moles using its molar mass:
molar mass of NaN3 = 65.01 g/mol
moles of NaN3 = mass / molar mass
moles of NaN3 = 65.0 g / 65.01 g/mol
moles of NaN3 = 0.999 moles
3. Use the balanced equation to determine the moles of N2 produced:
From the balanced equation, we see that 2 moles of NaN3 produce 3 moles of N2. Therefore, for 0.999 moles of NaN3, we can calculate the moles of N2 produced using the ratio:
moles of N2 = (0.999 moles NaN3) x (3 moles N2 / 2 moles NaN3)
moles of N2 = 1.4985 moles
4. Use the ideal gas law to calculate the volume of N2 gas:
PV = nRT
Where:
P = pressure (1.30 atm)
V = volume (to be calculated)
n = moles (1.4985 moles)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (converted from Celsius: °C + 273.15)
T = 88.0°C + 273.15 = 361.15 K
Rearranging the equation:
V = (nRT) / P
V = (1.4985 moles)(0.0821 L·atm/(mol·K))(361.15 K) / 1.30 atm
V = 37.77 L
Therefore, the volume of N2 gas produced is 37.77 L.
from the equation, you get the 3/2 moles of nitrogen as you had of sodium azide.
MolesN2=65g/molmassNaN3
then volume N2, get from PV=nRT