Asked by Argh
If 8.00 moles of NH3 reacted with 14.0 moles of O2, how many moles of H2O will be produced?
4NH3 (g) + 7O2 (g) --> 4NO2 + 6H2O (g)
4NH3 (g) + 7O2 (g) --> 4NO2 + 6H2O (g)
Answers
Answered by
DrBob222
How much H2O could you get if you used 8.00 moles NH3 and all of the oxygen needed. That will give you
8.00 x (6 moles H2O/4 moles NH3) = about 12.
Now how much H2O would be produced if you used 14.0 moles Oxygen and all of the NH3 needed. That will be 14.0 moles O2 x (6 moles H2O/7 moles O2) = 12
aha!. They are the same so you would get 12 moles H2O. What if they weren't the same. In that case you have a limiting reagent problem, the SMALLER value is the one you choose and the reagent producing the smaller value is the limiting reagent.
8.00 x (6 moles H2O/4 moles NH3) = about 12.
Now how much H2O would be produced if you used 14.0 moles Oxygen and all of the NH3 needed. That will be 14.0 moles O2 x (6 moles H2O/7 moles O2) = 12
aha!. They are the same so you would get 12 moles H2O. What if they weren't the same. In that case you have a limiting reagent problem, the SMALLER value is the one you choose and the reagent producing the smaller value is the limiting reagent.
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