Bài 1 : Chuỗi phản ứng :
a) \(Al\xrightarrow[]{\left(1\right)}Al_2O_3\xrightarrow[]{\left(2\right)}AlCl_3\xrightarrow[]{\left(3\right)}Al\left(OH\right)_3\)
b) \(Al\xrightarrow[\left(1\right)]{}Al_2O_3\xrightarrow[]{\left(2\right)}AlCl_3\xrightarrow[\left(3\right)]{}Al\left(OH\right)_3Al_2O_3\)
c) \(Fe\xrightarrow[]{\left(1\right)}FeSO_4\xrightarrow[]{\left(2\right)}FeCl_2\xrightarrow[]{\left(3\right)}Fe\left(OH\right)_2\xrightarrow[]{\left(4\right)}FeO\)
d) \(Zn\xrightarrow[]{\left(1\right)}ZnSO_4\xrightarrow[]{\left(2\right)}ZnCl_2\xrightarrow[]{\left(3\right)}Zn\left(OH\right)_2\xrightarrow[]{\left(4\right)}ZnO\)
e) \(Mg\left(OH\right)_2\xrightarrow[]{\left(1\right)}MgCl_2\xrightarrow[]{\left(2\right)}Mg\left(NO_3\right)_2\xrightarrow[]{\left(3\right)}Mg\left(OH\right)_2\xrightarrow[]{\left(4\right)}MgO\)
f) \(Fe\left(OH\right)_2\xrightarrow[]{\left(1\right)}FeO\xrightarrow[]{\left(2\right)}FeSO_4\xrightarrow[]{\left(3\right)}FeCl_2\xrightarrow[]{\left(4\right)}Fe\left(OH\right)_2\)
g) \(Fe\xrightarrow[]{\left(1\right)}FeCl_2\xrightarrow[]{\left(2\right)}Fe\left(NO_3\right)_2\xrightarrow[]{\left(3\right)}Fe\left(OH\right)_2\xrightarrow[]{\left(4\right)}FeSO_4\)
h) \(S\xrightarrow[]{\left(1\right)}SO_2\xrightarrow[]{\left(2\right)}SO_3\xrightarrow[]{\left(3\right)}H_2SO_4\xrightarrow[]{\left(4\right)}SO_2\)
k) \(Cu\xrightarrow[]{\left(1\right)}CuO\xrightarrow[]{\left(2\right)}CuSO_4\xrightarrow[]{\left(3\right)}Cu\left(NO_3\right)_2\xrightarrow[]{\left(4\right)}Cu\left(OH\right)_2\)
Bài 2 : Phân biệt các chất rắn
a) Hai chất rắn: \(Cao\) và \(P_2O_5\)
b) Hai chất rắn: BaO và \(P_2O_5\)
c) Hai chất rắn :\(Na_2O\) và \(P_2O_5\)
d) Hai chất rắn :\(K_2O\) và \(P_2O_5\)
e) Ba chất rắn : MgO, \(Na_2O\), \(P_2O_5\)
GIÚP MINH VỚI Ạ !!!!
đường glucozơ 
a, m tinh bột + m nước = m glucozo
m glucozo = m rượu etylic+ mCO2
b,
m tinh bột= 100.81%= 81 kg
=> m glucozo= 81+9= 90 kg
=> m rượu etylic= 90-44= 46 kg