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?

Rxn: $4 N {H}_{3} \left(a q\right) + 5 {O}_{2} \left(g\right) \rightarrow 4 N O \left(g\right) + 6 {H}_{2} O \left(l\right)$
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 $N {O}_{2}$. Can you write this redox reaction?
As to your question, we start with $\frac{15.8 \cdot g}{17.0 \cdot g \cdot m o {l}^{-} 1} = 0.929$ $m o l$ of ammonia. We get $\frac{21.8 \cdot g}{30 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.727$ $m o l$. Because it's 1 ammonia to 1 nitrous oxide, the yield is simply, (0.727*mol)/(0.929*mol) xx 100% = ??