In the Haber process, `30 L` of dhyrgen and `30L` of dintrogen were taken for reaction which yielded only `50%` of the expectedf product. What will be the xomposition of the gaseous mixturre under the aforesaid condition in the end?

Options

A. 20 L `NH_(3)`, 25 L `N_(2)` and 20 L `H_(2)`
B. 10 L `NH_(3), 25 `L `N_(2)` and 15 L `H_(2)`
C. 20 L `NH_(3)`, 10 L `N_(2)` and 30 L `H_(2)`
D. 20 L `NH_(3)`, 25 L `N_(2)` and 15 L `H_(2)`

Answer

B

Explanation

`N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)`
1 L of `N_(2)` reacts with 3 L of `H_(2)` to form 2 L of `NH_(3)`. Thus, `H_(2)` is limiting reagent.
10 L of `N_(2)` will react with 30 L of `H_(2)` to form 20 L of `NH_(3)`
Since actual yield is 50% of the expected value,
`NH_(3)` formed = 10 L
`N_(2)` reacted = 5 L
`N_(2)` unreacted = 30-5 = 25 L
`H_(2)` reacted = 15 L
`H_(2)` unreacted = 30-15 = 15 L
`therefore` Mixture will contains 10 L `NH_(3).25 L, N_(2), 15 L H_(2)`.