A mixture of NH3( g ) and N2H4( g ) is placed in a sealed container at 310 K . T

Question

A mixture of NH3(g) and N2H4(g) is placed in a sealed container at 310 K . The total pressure is 0.48 atm . The container is heated to 1200 K at which time both substances decompose completely according to the equations 2NH3(g)N2(g)+3H2(g)  ; N2H4(g)N2(g)+2H2(g) . After decomposition is complete the total pressure at 1200 K  is found to be 4.5 atm.

Find the percent amount (moles) of N2H4(g) in the original mixture. (Assume two significant figures for the temperature.)

Express your answer using two significant figures.

Explanation / Answer

Total pressure = 0.48 , T of container = 1200 K

Calculation of moles of both gases :

We use pV = nRT

Here p is pressure , V is volume , R is gas constant T is temperature in K

n is number of moles

value of R = 0.08206 L atm per (K mol)

T = 310 K

P = 0.48 atm

Lets plug all these values to get total moles

Total moles n ( n N2H4 + n NH3 ) = pV/RT

= [0.48 x 1.0 / (0.08206 x 310)] mol

= 0.018869 mol

(we assumed volume is 1.0 L )

After reaction is complete there are only moles of N2 and H2

So lets find total moles of N2 and H2

n (H2 + N2 ) = [4.5 x 1.0 / (0.08206 x 1200)] mol

= 0.04569 mol

In total moles Lets find mole of N2

By using reaction we say that when there are total 7 moles ( 2 N2 + 5 H2)

Then mol fraction of N2 = 2 / 7 = 0.28 lets use this mol fraction to get moles of N2 in 0.04569 mol

Mol of N2 = 0.04569 mol x 2 mol N2 / 7 mol

= 0.01306 mol N2

From reaction we know that half of N2 we get from 2 mol NH3 and half we get fom 1 mol of N2H4 so lets find half of moles of N2

= 1 /2 x 0.01306

= 0.00653 mol N2

And when 1 mol N2H4 decompose then we get 1 mol N2

So the mole ratio of N2 is 1 : 1

Moles of N2H4 = 0.00653 mol N2 x 1 mol N2H4 / 1 mol N2

= 0.00653 mol N2H4

Percent amount = mol of N2H4 / total moles f

= 0.00653 / 0.018869 x 100

=0.346 x 100

= 34.6 %

So the percent of N2H4 is 34.6 %   (mol percent

We can convert it to 2 sig fig

= 35 %

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