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

50.0 kg of `N_(2)` (g) and 10.0 kg of `H_(2)` (g) are mixed to produce `NH_(3)` (g). Calculate the amount of` NH_(3)` (g) formed. Identify the limiting reagent in the production of NH3 in this situation

Text Solution

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According to the balanced chemical reaction, `25 kg` of `N_(2)` reacts with `6 kg` of `H_(2)` to give `34 kg` of `NH_(3)`.
Thus, `6 kg` of `H_(2)` requires `-= 25 kg of N_(2)`
`10 kg` of `H_(2)` requries `-= (28 xx 10)/(6) = 46.6 kg or N_(2)`
Hence, `H_(2)` is the limiting reagent, since it is completely consumed in the reaction, and `H_(2)` will decied the amount of product formd.
Thus, `6 kg of H_(2) -= 34 kg of NH_(3)` ltbgt `10 kg of H_(2) -= (34 xx 10)/(6) = 56.6 kg of NH_(3)`
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