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`        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

```							Assume frst n1 and n2 be the initial moles of the two compunds,Hence initially total moles = n1+n2and from ideal gas eqn. we get,0.5*V = (n1+n2)*R*300after decmposition,from 2 moles of NH3 we get total 4 molesand from 1 moles of N2H4 we get 3 moles total.Hence the total moles for mixture => 2n1 + 3n2Anf from the ideal gas eqn. we get,4.5*V = (2n1 + 3n2)*R*1200Dividing both the eqns. we get,n2/(n1+n2) = 1/4and therfore,n1 = 3 and n2 = 1Percentage of N2H4 therfore is => 1*100/4 = 25%and % of  NH3 => 75%
```
2 years ago
```							assume n1 and n2 as the number of moles of nh3 and n2h4 initiallythey 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+n2consider the volume of the mixture initially as Vthe 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 equation0.5*V=[n1+n2]*R*300After they get decomposed 2 moles of nh3 gives 4 moles1 mole of n2h4 gives 3 moles, n2 moles of n2h4 would give 3n2 moles1 mole of nh3 gives 2 moles, n1 moles of nh3 would give 2n1 molesso therefore total moles after the decomposition is 2n1+3n2final pressure is 4.5 atmtemperature is 1200K4.5*V=[2n1+3n2]*R*1200after solving we get n1=3 and n2=1percentage of composition of nh3 is 75and that of n2h4 would be 25
```
7 months ago
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