Question

A mixture of NH3 and N2H4 is placed in a sealed container at 300K. The total pressure is 0.50 atm. The container is heated to 1200 K at which time both substances decompose entirely according to the equation 2NH3 -> N2 3H2, N2H4 -> N2 2H2. The total pressure at 1200 K is 4.5 atm. Find the percent of N2H4 in the original mixture.

Answer 1

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 %