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### Limiting reagent of NH(3) +CuO------->N(2)+H2O+CO

Posted: Sat Apr 07, 2018 4:56 pm
I understand that In this problem I am to balance the chemical equation, NH(3) +CuO------->N(2)+H2O+CO, which becomes 2NH(3)+3CuO--------->N2 +3H2O+3Cu, after that I have to find the moles of NH3(2.9 mol) and CuO (1.13 mol) however, from here I know I am to find the moles needed for each but, I am not quite sure how to achieve the answer to find the limiting reagent.

### Re: Limiting reagent of NH(3) +CuO------->N(2)+H2O+CO

Posted: Sun Apr 08, 2018 3:36 pm
You need to look at the balanced equation and look at which reactant will run out first based on the stoichiometry. One way to look is what is the maximum possible amount of product (pick 1) you can form if you have this much of 1 reactant and excess of others and pick the one that gives you the least amount. Another way to do this is to divide the starting moles by respective stoichiometry coefficient and pick the smallest.

### Re: Limiting reagent of NH(3) +CuO------->N(2)+H2O+CO

Posted: Sun Apr 15, 2018 3:37 pm
Steps to remember:

-Balance the chemical equation for the chemical reaction.
-Convert the given information into moles.
-Use stoichiometry for each individual reactant to find the mass of product produced.
-The reactant that produces a lesser amount of product is the limiting reagent.
-The reactant that produces a larger amount of product is the excess reagent.
-To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.

https://chem.libretexts.org/Core/Inorga ... g_Reagents

### Re: Limiting reagent of NH(3) +CuO------->N(2)+H2O+CO

Posted: Sun Jun 10, 2018 10:59 pm
Once you make your conversions, the smaller result is the limiting reagent and the larger one is excess. Subtract excess reagent by the overall excess reagent given