4K+O2=2K2O, n(K2O)=7.36/94=0.078 mol, n(K)=6.92/39=0.177 mol, n(O2)=4.28/32=0.133 mol, 0.078*2/0.177*100%=88.14%, 0.078/2*0.133*100%=29.32%, so the Percent Yield is 29.32%.
What Is The Percent Yield For The Reaction Of 6.92 G Of K And 4.28 G Of O2 If 7.36 G Of KO2 Is Recovered In The Laboratory?