2 NH ClO (s) ---> 4 H O(g) + Cl (g) + 2 O (g) + N (g)
In an experiment, 4.56 g NH ClO decomposes completely in a rigid 5.00-L container. What is the total pressure in the container if the
final temperature is 229 C?
Give your answer in atmospheres, accurate to three significant figures. Do not include the units as part of your answer.
Molar masses:
(in g mol )
NH ClO , 117.492
H O, 18.016
O , 32.00
N , 28.02
Cl , 70.90
I've been struggling with this question and I really need help :(
To solve this question, we need to use the ideal gas law equation, which is:
PV = nRT
where:
P is the pressure in atm
V is the volume in liters
n is the number of moles of gas
R is the ideal gas constant (0.0821 L*atm/mol*K)
T is the temperature in Kelvin
First, let's calculate the number of moles of NH ClO using the given mass and molar mass:
moles of NH ClO = mass / molar mass = 4.56 g / 117.492 g/mol = 0.0388 mol
Since the reaction is complete, the number of moles of other gases (H O, O , N , and Cl ) will be twice the number of moles of NH ClO:
moles of H O = 2 × 0.0388 mol = 0.0776 mol
moles of O = 2 × 0.0388 mol = 0.0776 mol
moles of N = 2 × 0.0388 mol = 0.0776 mol
moles of Cl = 2 × 0.0388 mol = 0.0776 mol
Next, let's convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = 229 °C + 273.15 = 502.15 K
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
P × 5.00 L = (0.0388 mol + 0.0776 mol + 0.0776 mol + 0.0776 mol) × 0.0821 L*atm/mol*K × 502.15 K
Simplifying the equation:
5.00 P = 0.2716 × 0.0821 × 502.15
Solving for P:
P = (0.2716 × 0.0821 × 502.15) / 5.00
P = 11.878 atm
Therefore, the total pressure in the container is 11.9 atm (rounded to three significant figures).
Note: In this calculation, we assumed that the volume of the container remains constant and that the gases behave ideally.