2 NaN3(s) = 2 Na(l) + 3 N2(g)
How many grams of NaN3 are needed to produce 20L of N2(g) at 303K and 776mmHg?
In the chemical reaction used in automotive air-bag safety systems, nitrogen gas is produced by the decomposition of sodium azide as described in the reaction below.
NaN3(s) Na(l) + N2(g)
What mass of NaN3 must be decomposed to generate enough nitrogen gas at 25.0 oC and 0.980 atm to fill an airbag with a volume of 5.0 x 104 cm3
To find the number of grams of NaN3 needed to produce 20L of N2(g) at 303K and 776mmHg, we can use the ideal gas law.
The ideal gas law is represented by the equation:
PV = nRT
Where:
P is the pressure (in atm)
V is the volume (in L)
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(K·mol))
T is the temperature (in Kelvin)
Since we're given the volume (20L), temperature (303K), and pressure (776mmHg), we can use these values in the ideal gas law equation to find the number of moles of N2(g).
First, we need to convert the pressure from mmHg to atm. We know that 1 atm = 760 mmHg. So,
776 mmHg * (1 atm / 760 mmHg) = 1.02 atm
Next, we rearrange the ideal gas law equation to solve for n:
n = PV / RT
Substituting the given values:
n = (1.02 atm * 20L) / (0.0821 L·atm/(K·mol) * 303K)
n = 0.0688 mol
According to the balanced equation, the molar ratio between NaN3 and N2 is 2:3. So, we can calculate the number of moles of NaN3 required using the ratio:
moles of NaN3 = (0.0688 mol N2) * (2 mol NaN3 / 3 mol N2)
moles of NaN3 = 0.0459 mol
Finally, we calculate the mass of NaN3 using its molar mass. The molar mass of NaN3 is:
Na: 22.99 g/mol
N: 14.01 g/mol
Therefore, molar mass of NaN3 = (22.99 g/mol) + (3 * 14.01 g/mol) = 65.0 g/mol
Mass of NaN3 = (0.0459 mol) * (65.0 g/mol) = 2.98 g
So, approximately 2.98 grams of NaN3 are needed to produce 20L of N2(g) at 303K and 776mmHg.
To find out how many grams of NaN3 are needed to produce 20L of N2(g) at 303K and 776mmHg, we need to use the concept of stoichiometry and the ideal gas law.
Step 1: Write the balanced equation for the reaction:
2 NaN3(s) → 2 Na(l) + 3 N2(g)
Step 2: Calculate the number of moles of N2 using the ideal gas law:
PV = nRT
where P is the pressure (in atm), V is the volume (in L), n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature (in K).
Convert the pressure given in mmHg to atm:
1 atm = 760 mmHg
So, 776 mmHg = 776/760 atm
Plug in the values into the ideal gas law equation:
(776/760) atm * 20 L = n * 0.0821 L·atm/mol·K * 303 K
Solving for n, we get:
n = ((776/760) * 20) / (0.0821 * 303)
Step 3: Use the stoichiometry of the balanced equation to find the moles of NaN3 needed. From the balanced equation, we know that 3 moles of N2 are produced for every 2 moles of NaN3:
moles of NaN3 = (3/2) * moles of N2
Step 4: Convert moles of NaN3 to grams:
To convert moles to grams, we need to use the molar mass of NaN3, which can be found by adding up the atomic masses of the elements in NaN3.
The atomic masses are:
Na: 22.99 g/mol
N: 14.01 g/mol
Molar mass of NaN3 = (22.99 * 2) + 14.01
Multiply the moles of NaN3 calculated in Step 3 by the molar mass of NaN3 to get the grams of NaN3 needed.
That's how you calculate the number of grams of NaN3 needed to produce 20L of N2(g) at 303K and 776mmHg.
Use PV = nRT. Substitute and solve for n = number of mols.
Then use the coefficients in the balanced equation to convert mols N2 to mols NaN3.
Finally, g NaN3 = mols NaN3 x molar mass NaN3.