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Under S.T.P. 1 mol of N(2) and 3 mol of ...

Under S.T.P. 1 mol of `N_(2)` and 3 mol of `H_(2)` will form on complete reaction

A

4 moles of `NH_(3)`

B

89.6 L of `NH_(3)`

C

22.4 L of `NH_(3)`

D

44.8 L of `NH_(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`underset(1mol)(N_(2))+underset(3mol)(3H_(2))tounderset(2mol)(2NH_(3))`
1 mol of `N_(2)` reacts with 3 moles of `H_(2)` to give 2 moles of `NH_(3)` which occupies a volume
`= 2xx22.4 = 44.8 L` at S.T.P.
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