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Grade: 12
        A mixture of NH3 and N2H4 is placed in a sealed container at 300K. The total pressure is 0.5 atm. When container is heated to 1200K both substance decompose completely according to the equation 2NH3 ~> N2 +3 H2.   ;  N2H4~>N2+ 2 H2.     After decomposition, the total pressure at 1200K is found to be 4.5 atm.  What is the % of N2 H4 in the original mixture?
2 years ago

Answers : (2)

Vikas TU
9499 Points
							
Assume frst n1 and n2 be the initial moles of the two compunds,
Hence initially total moles = n1+n2
and from ideal gas eqn. we get,
0.5*V = (n1+n2)*R*300
after decmposition,
from 2 moles of NH3 we get total 4 moles
and from 1 moles of N2H4 we get 3 moles total.
Hence the total moles for mixture => 2n1 + 3n2
Anf from the ideal gas eqn. we get,
4.5*V = (2n1 + 3n2)*R*1200
Dividing both the eqns. we get,
n2/(n1+n2) = 1/4
and therfore,
n1 = 3 and n2 = 1
Percentage of N2H4 therfore is => 1*100/4 = 25%
and % of  NH3 => 75%
2 years ago
George Thomas
13 Points
							
assume n1 and n2 as the number of moles of nh3 and n2h4 initially
they have given initial pressure at 300 K as 0.5atm,
initial there is a mixture of n2h4 and nh3 so the total number of moles is n1+n2
consider the volume of the mixture initially as V
the volume after they get decomposed is also V .
the same question is there in cengage over there they mentioned that point. that the initial volume of mixture= final volume after geting decomposed here its not mentioned!
From ideal gas equation
0.5*V=[n1+n2]*R*300
After they get decomposed 
2 moles of nh3 gives 4 moles
1 mole of n2h4 gives 3 moles, n2 moles of n2h4 would give 3n2 moles
1 mole of nh3 gives 2 moles, n1 moles of nh3 would give 2n1 moles
so therefore total moles after the decomposition is 2n1+3n2
final pressure is 4.5 atm
temperature is 1200K
4.5*V=[2n1+3n2]*R*1200
after solving we get n1=3 and n2=1
percentage of composition of nh3 is 75
and that of n2h4 would be 25
 
 
 
7 months ago
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