PTHH
2Zn + O2 -to> 2ZnO
0.08....0.04..........0.08... mol
\(\dfrac{0.08}{2}< \dfrac{0.25}{1}O_2du,baitoantinhtheoZn\)
nZn = \(\dfrac{5.2}{65}\)= 0.08 mol
nO2 = \(\dfrac{5.6}{22.4}\) = 0.25 mol
A gồm O2 dư va ZnO
mO2(du) = (0.25-0.04)*32 =6.72 g
mZnO = 0.08*81=6.48 g
%mZnO = \(\dfrac{6.48}{6.48+6.72}\).100%=49.1%
%mO2=100% - 49.1%= 50.9%