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

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

1.	Under S.T.P. 1 mol of `N_(2)` and 3 mol of `H_(2)` will form on complete reactionA. 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)`
Answer» Correct Answer - 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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Under S.T.P. 1 mol of `N_(2)` and 3 mol of `H_(2)` will form on complete reactionA. 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)`

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