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9 October, 04:28

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.

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  1. 9 October, 04:44
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    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 %
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