20.0 kg of N(2)(g) and 3.0 kg of H(2)(g)...

20.0 kg of `N_(2)(g)` and 3.0 kg of `H_(2)(g)` are mixed to produce `NH_(3)(g)`. The amount of `NH_(3)(g)` formed is

17 kg

34 kg

20 kg

3 kg

Text Solution

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The correct Answer is:

`underset(28)(N_(2)) + underset(2 xx 3)(3H_(2)) to underset(34) (2NH_(3))`
1 mole of `H_(2)` can react with `28/6 xx 3= 17` kg

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