How would you determine the percent yield for the reaction between 15.8g of NH3 and excess oxygen to produce 21.8g of NO gas and water?

1 Answer
anor277
Nov 23, 2015

Rxn: #4NH_3(aq) + 5O_2(g) rarr 4NO(g) + 6H_2O(l)#

Explanation:

Is the reaction above a redox reaction? What is oxidized, and what is reduced? What are the oxidation states? Nitrous oxide is capable of being further oxidized to #NO_2#. Can you write this redox reaction?

As to your question, we start with #(15.8*g)/(17.0*g*mol^-1) = 0.929# #mol# of ammonia. We get #(21.8*g)/(30*g*mol^-1)# #=# #0.727# #mol#. Because it's 1 ammonia to 1 nitrous oxide, the yield is simply, #(0.727*mol)/(0.929*mol) xx 100% = ??#

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