Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.25 atm at a temperature of 20.0°C. How much sodium azide is needed if the air bag must produce the same pressure at 10.0°C?
20.0°C:
_______ g NaN3
(Work) below
n=PV
-----
RT
I used 25X25X20 And divided by 1000 which gave me 12.5 as
Advised for my L(Volume) then converted 20celsius to K which gave me 293.15K and my constant R .08206 so i think i have everything but my solution is wrong
(1.25atms) X (12.5L)
--------------------- =.65 moles (It says i'm wrong what am i doing?)
(0.08206 X 293.15K)
Are you sure you are using the same chemical equation I used; i.e., 2NaN3 ==> 2Na + 3N2. If so you haven't done anything wrong. 0.65 gives you the mols N2. Now you convert mols N2 to mols NaN3. That is 0.65 mols N2 x (2 mols NaN3/3 mols N2 = ?
Then grams NaN3 = mols NaN3 x molar mass NaN3.
To calculate the amount of sodium azide (NaN3) needed to inflate the bag, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, we need to convert the given dimensions of the bag from cm to liters. The volume will be (25.0 cm) x (25.0 cm) x (20.0 cm) = 12,500 cm^3 = 12.5 L.
Next, we need to convert the temperature from Celsius to Kelvin. The temperature in Kelvin will be 20.0°C + 273.15 = 293.15 K.
Now we can calculate the number of moles using the ideal gas law equation:
n = (P x V) / (R x T)
n = (1.25 atm x 12.5 L) / (0.08206 L.atm/mol.K x 293.15 K)
n ≈ 0.63 moles
Therefore, to inflate the bag to a pressure of 1.25 atm at a temperature of 20.0°C, approximately 0.63 moles of sodium azide (NaN3) are required.
Note: The molecular weight of NaN3 is approximately 65 grams/mol. To find the mass of NaN3 needed, multiply the number of moles by the molecular weight:
mass of NaN3 = 0.63 moles x 65 g/mol ≈ 40.95 g
So approximately 40.95 grams of sodium azide (NaN3) are needed for inflation at 20.0°C.
To calculate the amount of sodium azide (NaN3) needed to inflate the bag to a certain pressure and temperature, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles
R = gas constant (0.08206 L·atm/mol·K)
T = temperature in Kelvin (K)
For the first part of the question, where the temperature is 20.0°C, you need to convert it to Kelvin by adding 273.15:
T = 20.0°C + 273.15 = 293.15 K
Next, convert the volume of the bag to liters by dividing by 1000:
V = (25.0 cm) x (25.0 cm) x (20.0 cm) ÷ 1000 = 12.5 L
Now, substitute these values into the equation:
(1.25 atm) x (12.5 L) = n x (0.08206 L·atm/mol·K) x (293.15 K)
Solving for n (moles):
n = (1.25 atm x 12.5 L) / (0.08206 L·atm/mol·K x 293.15 K) ≈ 0.6485 moles
So, you would need approximately 0.6485 moles of NaN3 to inflate the bag to a pressure of 1.25 atm at a temperature of 20.0°C.
For the second part of the question, where the temperature is 10.0°C, follow the same steps as above, but substitute the new temperature value and recalculate:
T = 10.0°C + 273.15 = 283.15 K
V = 12.5 L (same as before)
(1.25 atm) x (12.5 L) = n x (0.08206 L·atm/mol·K) x (283.15 K)
Solving for n (moles):
n = (1.25 atm x 12.5 L) / (0.08206 L·atm/mol·K x 283.15 K) ≈ 0.6812 moles
Therefore, you would need approximately 0.6812 moles of NaN3 to produce the same pressure in the bag at a temperature of 10.0°C.