A car air bag requires 70. L of nitrogen gas to inflate properly. The following equation represents the production of nitrogen gas:
2NAN subscript 3 yielding 2NA + 3N subscript 2
the density of nitrogen gas is typically 1.16 g/L at room temperature. How would you calculate the number of grams of Nitrogen that is needed to inflate the air bag?
First let me point out that NAN3 is NOT the same as NaN3 and NA is not the same as Na.
2NaN3 ==> 2Na + 3N2
How about 1.16 g/L x 70 L = ?? g.
To calculate the number of grams of nitrogen needed to inflate the airbag, you can follow these steps:
1. Determine the number of moles of nitrogen gas needed. We know that the airbag requires 70 L of nitrogen gas, and the equation shows that 3 moles of N₂ is produced for every 2 moles of NaN₃.
Moles of N₂ = (70 L N₂) x (3 mol N₂ / 2 mol NaN₃)
2. Calculate the mass of nitrogen gas using the molar mass of nitrogen. The molar mass of nitrogen (N₂) is 28.02 g/mol.
Mass of N₂ = Moles of N₂ x Molar mass of N₂
3. We can use the density of nitrogen gas to convert the mass of nitrogen to the volume.
Mass of N₂ = Volume of N₂ x Density of N₂
Density of N₂ = 1.16 g/L
Volume of N₂ = 70 L (given)
Mass of N₂ = 70 L x 1.16 g/L
Therefore, the number of grams of nitrogen needed to inflate the airbag can be calculated by multiplying the volume of nitrogen gas (70 L) with the density of nitrogen gas (1.16 g/L).
To calculate the number of grams of nitrogen needed to inflate the airbag, we need to determine the number of moles of nitrogen gas required and then convert that to grams.
Here's how you can calculate it step by step:
Step 1: Convert the given volume of nitrogen gas (70 L) to moles.
To do this, we can use the ideal gas law equation: PV = nRT, where
P = pressure (assumed to be constant for room temperature and pressure),
V = volume of the gas (70 L),
n = number of moles of gas,
R = ideal gas constant (0.0821 L·atm/(mol·K)),
T = temperature (room temperature, typically around 298 K).
Rearranging the equation, we can solve for moles (n):
n = PV / RT
Here, pressure (P) and temperature (T) are not given, but since we're assuming room temperature and pressure, we can use standard values:
P = 1 atm (atmospheric pressure)
T = 298 K
Substituting the values into the equation:
n = (1 atm) * (70 L) / (0.0821 L·atm/(mol·K) * 298 K)
n ≈ 2.38 moles
Step 2: Use stoichiometry to determine the moles of nitrogen gas in the chemical equation.
From the given equation, it is known that 2 moles of NaN₃ yield 3 moles of N₂.
Therefore, the ratio of moles of nitrogen gas to moles of NaN₃ is 3/2.
So, moles of nitrogen gas = (2.38 moles NaN₃) * (3 moles N₂ / 2 moles NaN₃)
moles of nitrogen gas ≈ 3.57 moles
Step 3: Convert moles of nitrogen gas to grams.
To do this, we need to use the molar mass of nitrogen gas (N₂), which is approximately 28.02 g/mol.
mass of nitrogen gas = (3.57 moles) * (28.02 g/mol)
mass of nitrogen gas ≈ 99.98 grams
Therefore, approximately 100 grams of nitrogen gas is needed to inflate the airbag properly.