In Haber process 30 litre of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only`50%` of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end ?

Options

A. `20 L` ammonia, `25 L` nitogen, `15 L` hydrogen
B. `20 L` ammonia, `10 L` nitogen, `30 L` hydrogen
C. `10 L` ammonia, `25 L` nitogen, `15 L` hydrogen
D. `20 L` ammonia, `20 L` nitogen, `20 L` hydrogen

Answer

C

Explanation

According to the balanced equation,
`{:(N_(2)(g)+,3H_(2)(g)rarr,2NH_(3)(g),,),("1 mol","3 mol","2 mol",,),("1 vol","3 vol","2 vol",,"(Avogadro's law)"):}`
`30 L N_(2)` will produce `60 L NH_(3)` while `30L H_(2)` will produce `20L NH_(3)`. Thus `H_(2)` is the limiting reacant and `20L NH_(3)` will be formed. As we get get only `50%` of the expected product, `10L NH_(3)` is produced. Hence,
? vol `H_(2) = 10L NH_(3) xx (3L H_(2))/(2LNH_(3)) = 15 L H_(2)` consumed
? vol `N_(2) = 10 L NH_(3) xx (1 L N_(2))/(2L NH_(3)) = 5L N_(2)` consumed
Thus, we are left with `25L N_(2)` and `15L H_(2)`