\(2Al_2O_3\xrightarrow[Criolit]{đpnc}4Al+3O_2\)
\(2Al+3S\xrightarrow[]{t^o}Al_2S_3\)
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`AlCl_3 + 3KOH -> Al(OH)_3 + 3KCl`
\(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)
`Al_2O_3 + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2O`
`Al_2(SO_4)_3 + 3BaCl_2 -> 2AlCl_3 + 3BaSO_4`
al → al2o3 → al2(so4)3 → alcl3 → al(oh)3 → al(no3)3 → al → al2(so4)3 → al(oh)3 → al2o3 → al →naalo2
Al --> Al2O3 --> AlCl3 --> Al2(SO4)3 --> Al(OH)3 --> Al2O3 --> Al
hoàn thành sơ đồ sau :
AL ---> Al2O3 ---> AlCl3 ---> Al(OH)3 ---> Al2(SO4)3
Viết PTHH
a, Al-->Al2O3-->AlCl3-->Al(OH)3-->Al203-->Al
"(AlCl3-->Al(NO3)3
b, Fe-->Fe3O4-->FeCl3-->Fe(OH)3-->Fe2O3-->Fe
c, Fe(OH)3-->Fe2(SO4)3-->FeCl3-->Fe(NO3)3
AL--->AL2O3--->ALCL3--->AL(OH)3--->AL2O3--->AL--->ALCL3
Giúp e vs ạ hoá lớp 9
Hoàn thành dãy biến hoá sau
Alcl3----->al(OH)
Al(OH)3 ---->NaAlo2
Al2o3 ------>NaAlo2
Al---->Al2O3---->AlCl3---->Al(OH)3 ---->NaAlO2
GIÚP MÌNH
hoàn thành sơ đồ phản ơngs sau giúp em với
Al -> Al2 -> AlCl3 -> Al (OH)3 -> Al2 (SO4)3
Them 0,6 mol NaOH vao dung dich chua x mol AlCl3 thu duoc 0,2 mol Al(OH)3.Them tiep 0,9 mol NaOH thay so mol cua Al(OH)3 la 0,5. Them tiep 1,2 mol NaOH nua thay so mol Al(OH)3 van la 0,5 mol. Tinh x?
