2NaN(3) > 2 Na +3N(2)
What mass of sodium azide must be reacted to inflate and air bag to 70.0 L at STP?
mols 70.0L x (22.4L/mol) = ? mols N2
Convert mols N2 to mols NaN3 using the coefficients in the balanced equation.
Now convert mols NaN3 to g. g NaN3 = mols NaN3 x molar mass NaN3
To determine the mass of sodium azide required to inflate an airbag to a specific volume at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation:
PV = nRT,
where:
P = pressure (in atmospheres),
V = volume (in liters),
n = number of moles of gas,
R = ideal gas constant (0.0821 L·atm/mol·K),
T = temperature (in Kelvin).
Given:
V = 70.0 L (volume of the airbag)
P = 1 atm (pressure at STP)
T = 273 K (temperature at STP)
Using the ideal gas law equation, we can solve for the number of moles of gas (n):
n = PV / RT.
n = (1 atm * 70.0 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ (70.0 L) / (22.414 L/mol)
n ≈ 3.125 moles.
From the balanced chemical equation:
2 NaN3 → 2 Na + 3 N2,
we can see that 2 moles of NaN3 react to form 2 moles of Na.
Therefore, the number of moles of NaN3 required will be half the number of moles of Na:
n(NaN3) = n(Na) / 2
n(NaN3) ≈ 3.125 moles / 2
n(NaN3) ≈ 1.5625 moles.
Now, we need to convert moles of NaN3 to mass:
To do that, we need the molar mass of NaN3. By adding up the atomic masses of sodium (Na) (22.99 g/mol) and nitrogen (N) (14.01 g/mol), we get:
Molar mass of NaN3 = (22.99 g/mol) + (3 * 14.01 g/mol)
Molar mass of NaN3 ≈ 65.99 g/mol.
Finally, we can calculate the mass of NaN3 required using the moles and molar mass:
Mass of NaN3 = n(NaN3) * Molar mass of NaN3
Mass of NaN3 ≈ 1.5625 moles * 65.99 g/mol
Mass of NaN3 ≈ 103.18 g (rounded to two decimal places).
Therefore, approximately 103.18 grams of sodium azide must be reacted to inflate an airbag to 70.0 L at STP.
To answer this question, we need to follow a few steps:
Step 1: Balance the chemical equation.
The given equation is: 2NaN3(3) > 2Na + 3N2
Step 2: Calculate the molar mass of sodium azide.
The molar mass of sodium azide (NaN3) can be calculated by adding up the atomic masses of individual elements:
1 Na (Sodium) + 3 N (Nitrogen) = 22.99 + (3 * 14.01) = 65.02 g/mol
Step 3: Use the ideal gas law to calculate the number of moles of nitrogen gas (N2) required.
The ideal gas law formula is: PV = nRT
Where:
P = Pressure (STP = 1 atm)
V = Volume (70.0 L)
n = number of moles
R = gas constant (0.0821 L.atm/mol.K)
T = Temperature (STP = 273.15 K)
Rearranging the formula to solve for n:
n = PV / RT
Substituting the known values:
n = (1 atm) * (70.0 L) / (0.0821 L.atm/mol.K * 273.15 K)
Step 4: Calculate the number of moles of sodium azide required.
From the balanced equation, we can see that the molar ratio between sodium azide (NaN3) and nitrogen gas (N2) is 2:3. So, for every 2 moles of NaN3, we get 3 moles of N2.
Using the calculated moles of N2 from step 3, we can set up a proportion:
(2 moles NaN3) / (3 moles N2) = (x moles NaN3) / (moles calculated in step 3)
Cross-multiply and solve for x:
(2 / 3) = (x / moles calculated in step 3)
Step 5: Calculate the mass of sodium azide required.
Using the molar mass of sodium azide from step 2, we can convert moles of NaN3 to grams:
Mass = moles of NaN3 * molar mass of NaN
Substituting the calculated moles of NaN3 from step 4:
Mass = (x moles NaN3) * (molar mass of NaN3)
Finally, calculate the mass of sodium azide required using the above formula.