Question

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?

Answer 1

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% = ??`