Answer:
The percent of N2H4 in the original mixture is 25 %
Explanation:
Step 1: Data given
Temperature in the sealed container = 300K
The total pressure = 0.50 atm
The container is heated to 1200K
The total pressure at 1200K = 4.5 atm
Step 2: The balanced equation
2NH3 → N2 + 3H2
N2H4 → N2 + 2H2
Step 3: Calculate the initial moles
p*V = n*R*T
⇒ with p = the total pressure = 0.50 atm
⇒ with V = the volume of the gas
⇒ with n = the total moles of the gasses (n1 + n2)
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = the temperature in the container = 300
0.50*V= (n1+n2)*R*300
Step 4: after decmposition,
from 2 moles of NH3 we'll get 4 moles (n1 → 2n1)
from 1 moles of N2H4 we'll get 3 moles. (n2 → 3n2)
The total moles for mixture = 2n1 + 3n2
p*V= n*R*T
⇒ with p = the total pressure at 1200 K = 4.5 atm
⇒ V = The volume
⇒ with n = the number of moles
⇒ with R = the gas constant = 0.08206 L* atm/ K*mol
⇒ with T = the temperature = 1200 K
4.5*V = (2n1 + 3n2)*R*1200
0.50*V= (n1+n2)*R*300
Step 5: Calculate the percentage of N2H4
After dividing both equations we get:
n2/(n1+n2) = 1/4
n1 = 3 and n2 = 1
Percentage of N2H4 therfore is => 1*100/4 = 25%
and % of NH3 => 75%
The percent of N2H4 in the original mixture is 25 %
