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50.0 kg of N(2)(g) and 10.0 kg of H(2)(g...

50.0 kg of `N_(2)(g)` and 10.0 kg of `H_(2)(g)` are mixed to produce `NH_(3)(g)`. Calculate the `NH_(3)(g)` formed. Identify the limiting reagent in the production of `NH_(3)` in this situation.

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`3.30 xx 10^(3)` mol `NH_(3)` is obtained.
Dihydrogen is the limiting agent.
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