Apply the gas laws to determine how many grams of sodium azide (NaN3) are needed to fill the air bag of your vehicle to a volume of 79.0 L at 35.0 degrees celsius and 1.00 atm. The reaction occurring in the air bag can be repesented by the equation, 2NaN3 (s)-> 2Na (s) + 3 N2 (g). The gas constant, R, is equal to 0.0821 L atm mol ^-1 K ^-1
2NaN3 ==> 2Na + 3N2
Use PV = nRT to convert 79.0 L N2 to moles at the conditions listed.
Then use the coefficients in the balanced equation to convert moles N2 to moles NaN3. Finally, moles NaN3 = grams NaN3/molar mass NaN3.
To determine the number of grams of sodium azide (NaN3) needed to fill the airbag of your vehicle, we can use the ideal gas law equation: PV = nRT.
Given:
- Volume (V) = 79.0 L
- Temperature (T) = 35.0 degrees Celsius = 273.15 + 35 = 308.15 K
- Pressure (P) = 1.00 atm
- Gas constant (R) = 0.0821 L atm mol^-1 K^-1
Step 1: Convert temperature to Kelvin (K)
Temperature in Kelvin can be obtained by adding 273.15 to the Celsius value.
308.15 K
Step 2: Rearrange the ideal gas law equation to solve for the number of moles (n)
PV = nRT => n = PV / RT
Step 3: Plug in the values into the equation and calculate the number of moles
n = (1.00 atm) * (79.0 L) / (0.0821 L atm mol^-1 K^-1 * 308.15 K)
n = 2.67 mol (rounded to two decimal places)
Step 4: Use stoichiometry to convert moles of NaN3 to grams
From the balanced equation, 2 mol of NaN3 produce 3 mol of N2.
To find the grams of NaN3, we need to convert moles to grams using the molar mass.
The molar mass of NaN3 can be calculated as:
Na = 22.99 g/mol (sodium)
N = 14.01 g/mol (nitrogen)
Therefore, molar mass of NaN3 = (22.99 x 2) + 14.01 = 65.99 g/mol
To convert moles to grams, we can use the formula:
grams = moles * molar mass
grams = 2.67 mol * 65.99 g/mol
grams = 176.29 g (rounded to two decimal places)
Therefore, approximately 176.29 grams of sodium azide (NaN3) are needed to fill the airbag of your vehicle to a volume of 79.0 L at 35.0 degrees Celsius and 1.00 atm.