Sodium azide (NaN3) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation:
2NaN3 --> 2Na + 3N2
If the Nitrogen gas (N2) gas that results from the decomposition of a 5.00g sample of NaN3 could be collected by displacing water from an inverted flask, what volume of gas would be produced at 21⁰C and 762mmHg?
To calculate the volume of gas produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, we need to calculate the number of moles of N2 gas produced by the decomposition of NaN3.
Given:
Mass of NaN3 = 5.00g
The molar mass of NaN3 is:
Na = 22.99 g/mol
N = 14.01 g/mol
So, the molar mass of NaN3 is:
3(14.01) + 22.99 = 65.00 g/mol
To calculate the number of moles of NaN3:
moles of NaN3 = mass / molar mass
moles of NaN3 = 5.00g / 65.00 g/mol = 0.0769 mol
According to the balanced chemical equation:
2 mol of NaN3 produce 3 mol of N2 gas.
So, the number of moles of N2 gas produced is:
moles of N2 = (0.0769 mol / 2 mol of NaN3) * 3 mol of N2 = 0.1154 mol
Now we can calculate the volume of the gas produced using the ideal gas law equation.
Given:
Temperature (T) = 21⁰C = 21 + 273 = 294 K
Pressure (P) = 762 mmHg
Converting the pressure to atmospheres:
1 atm = 760 mmHg
Pressure (P) = 762 mmHg / 760 mmHg/atm = 1.0026 atm
Using the ideal gas law equation:
PV = nRT
V = nRT / P
V = (0.1154 mol * 0.0821 L.atm/mol.K * 294 K) / 1.0026 atm
V ≈ 3.378 L
Therefore, approximately 3.378 liters of N2 gas would be produced.