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 of ammonia, 25 L of nitrogen, 15 L of hydrogen
B. 20 L of ammonia, 20L of nitrogen, 20 L hydrogen
C. 10L of ammonia, 25 L of nitrogen, 15 L of hydrogen
D. 20L of ammonia, 10 L of nitrogen, 30 L of hydrogen

Answer

C

Explanation

`underset("1vol")(N_(2))+underset("3vol")(3H_(2))hArr underset("2vol")(2NH_(3))`
`:.` 30 L of `H_(2)` will react with 10 L of `N_(2)` to produce 20 L of `NH_(3)` (Here `H_(2)` is the limiting reactant). As the yield is only 50%, 15 L of `H_(2)` will react with 5 L of `N_(2)` to produce 10 L of `NH_(3)`
`:.` Reaction mixture will contain,
15 L of `H_(2)` (30L-15L), and 10 L of `NH_(3)`