what volume of nitrogen gas is produced at 1.00 atm and 27C when 101 g of NaN3 decomposes
To determine the volume of nitrogen gas produced when 101 g of NaN3 decomposes at 1.00 atm and 27°C, we need to use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm / (mol·K))
T = temperature (in Kelvin)
First, let's calculate the number of moles of nitrogen gas produced. We can use the molar mass of NaN3 and the given mass of NaN3 to find the number of moles.
1. Calculate the molar mass of NaN3:
The molar mass of Na is 22.99 g/mol, and the molar mass of N is 14.01 g/mol.
Molar mass of NaN3 = (22.99 g/mol) + (3 × 14.01 g/mol) = 65.00 g/mol
2. Calculate the number of moles of NaN3:
Number of moles = mass / molar mass
Number of moles = 101 g / 65.00 g/mol = 1.55 mol
Now let's convert the temperature to Kelvin:
Temperature in Kelvin = 27°C + 273.15 = 300.15 K
Now, substitute the values into the ideal gas law equation:
PV = nRT
(1.00 atm) * V = (1.55 mol) * (0.0821 L·atm / (mol·K)) * (300.15 K)
V = (1.55 mol * 0.0821 L·atm / (mol·K) * 300.15 K) / (1.00 atm)
V = 38.6 L
Therefore, the volume of nitrogen gas produced is 38.6 liters at 1.00 atm and 27°C when 101 g of NaN3 decomposes.