1)
2Fe(OH)3-->Fe2O3+3H2O
3Ba(NO3)2+Al2(SO4)3-->3BaSO4+2Al(NO3)3
3H2SO4+2Al-->Al2(SO4)3+3H2
P2O5+3H2O-->2H3PO4
2)
дано
m(CuSO4)=80 g
NaOH
----------------------
m(Cu(OH)2)-?
80 g Xg
CuSO4+2NaOH-->Na2SO4+Cu(OH)2
160 98
M(CuSO4)=160 g/mol , M(Cu(OH)2)=98 g/mol
80/160 = X/98
X=49g
ответ 49 г