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Topic: limiting reagents (pls help me asap)  (Read 3492 times)

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superman87

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limiting reagents (pls help me asap)
« on: February 25, 2005, 04:42:31 AM »
i dont understand how to do limiting reagents calculations if the heat is given

ex - N2 + 3H2 = 2NH3 +46.19 Kj


if 3 mols of N2 react with 7 mols of H2 how much energy will be given off? how much excess reagent is left over?

can any1 show me how to do this problem step by step. pls pls pls

Offline AWK

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Re:limiting reagents (pls help me asap)
« Reply #1 on: February 25, 2005, 05:54:49 AM »
You should find, which reagent is in excess. (fing the factor for the reaction:
N2 + 3H2 = 2NH3 +46.19 Kj
which give you 3N2 or 7H2 on the left side). The lower number will be also a multiplier for the energy of the above reaction.
The rest of your problem is strightforward, I do hope.
« Last Edit: February 25, 2005, 08:30:48 AM by AWK »
AWK

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