The ¡°air¡± that fills the air-bags installed in automobiles is actually nitrogen gas produced by the
decomposition of sodium azide, NaN3. Assuming that the decomposition of sodium azide produces both
nitrogen and sodium metal (Na), what volume, in litres, of nitrogen gas is released from the
decomposition of 1.88 g of sodium azide? The pressure is 755 torr and the temperature is 24oC.
Complete this question showing all of your work
Write and balance the equation.
Convert 1.88 g NaN3 to mols. mol = grams/molar mass.
Using the coefficients in the balanced equation, convert mols NaN3 to mols N2.
Now convert mols N2 at the conditions listed by using PV = nRT and solving for L. Remember P must be in atm and T in kelvin.
Answer is 1.104L
Answer is 1.104L Oh remember that there are 0.028 mol in 1.88g of NaN2. Final, in 0.028mol of NaN3, there is 0.042 mol of N2. This is your n.
so, V= ( nRT)/P remember, you are given R, T and P
Hope this helps:)
To solve this question, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles
R is the ideal gas constant (0.0821 L*atm/mol*K)
T is the temperature (in Kelvin)
However, the given values for pressure and temperature are in torr and degrees Celsius, respectively. Therefore, we need to convert them to the appropriate units before proceeding.
First, let's convert the pressure from torr to atm:
1 atm = 760 torr
So, the pressure in atm = 755 torr / 760 torr/atm ≈ 0.9934 atm.
Next, we need to convert the temperature from degrees Celsius to Kelvin:
Kelvin = Celsius + 273.15
Therefore, the temperature in Kelvin = 24°C + 273.15 ≈ 297.15 K.
Now, we can rearrange the ideal gas law equation to solve for the volume:
V = (nRT) / P
To find the number of moles (n) of nitrogen gas, we need to determine the molar mass of sodium azide (NaN3).
Molar mass of NaN3 = (22.99 g/mol (Na) + 14.01 g/mol (N) x 3) = 65.01 g/mol
To calculate the number of moles, we will use the equation:
moles = mass / molar mass
moles of NaN3 = 1.88 g / 65.01 g/mol ≈ 0.0289 mol
Now, let's calculate the volume of nitrogen gas:
V = (0.0289 mol x 0.0821 L*atm/mol*K x 297.15 K) / 0.9934 atm
V ≈ 0.686 L
Therefore, approximately 0.686 liters of nitrogen gas is released from the decomposition of 1.88 g of sodium azide.