Câu 1 ) A / Mg(NO3) \(\underrightarrow{\left(1\right)}\) Mg(OH)2 \(\underrightarrow{\left(2\right)}\) MgCl2\(\underrightarrow{\left(3\right)}\) KCl \(\underrightarrow{\left(4\right)}\) KNO3
B/ \(Na\underrightarrow{\left(1\right)}Na_2O\underrightarrow{\left(2\right)}NaOH\underrightarrow{\left(3\right)}Na_2SO_4\underrightarrow{\left(4\right)}NaCl\underrightarrow{\left(5\right)}NaNO_3\underrightarrow{\left(6\right)}NaCl\underrightarrow{\left(7\right)}NaOH\)
C/\(Mg\underrightarrow{\left(1\right)}MgO\underrightarrow{\left(2\right)}MgCl_2\underrightarrow{\left(3\right)}Mg\left(NO_3\right)_2\underrightarrow{\left(4\right)}Mg\left(OH\right)_2\underrightarrow{\left(5\right)}MgSO_4\underrightarrow{\left(6\right)}MgCO_3\)
D/\(CuSO_4\underrightarrow{\left(1\right)}Cu\left(OH\right)_2\underrightarrow{\left(2\right)}CuO\underrightarrow{\left(3\right)}CuCl_2\underrightarrow{\left(4\right)}Cu\left(OH\right)_2\underrightarrow{\left(5\right)}CuSO_4\)
E/\(CuCl_2\underrightarrow{\left(1\right)}Cu\left(OH\right)_2\underrightarrow{\left(2\right)}CuSO_4\underrightarrow{\left(3\right)}Cu\underrightarrow{\left(4\right)}CuO\)
F/ \(Cu\left(OH\right)_2\underrightarrow{\left(1\right)}CuO\underrightarrow{\left(2\right)}CuCl_2\underrightarrow{\left(3\right)}Cu\left(NO_3\right)_2\underrightarrow{\left(4\right)}NaNO_3\)
G/\(Fe_2O_3\underrightarrow{\left(1\right)}FeCl_3\underrightarrow{\left(2\right)}Fe\left(OH\right)_3\underrightarrow{\left(3\right)}Fe_2O_3\underrightarrow{\left(4\right)}Fe_2\left(SO_4\right)_3\)
H/ \(ZnCl_2\underrightarrow{\left(1\right)}Zn\left(OH\right)_2\underrightarrow{\left(2\right)}ZnCl_2\underrightarrow{\left(3\right)}NaCl\underrightarrow{\left(4\right)}NaNO_3\)
M/\(CuO\underrightarrow{\left(1\right)}CuCl_2\underrightarrow{\left(2\right)}Cu\left(OH\right)_2\underrightarrow{\left(3\right)}CuO\underrightarrow{\left(4\right)}CuSO_4\)
N/\(Fe\left(OH\right)_2\underrightarrow{\left(1\right)}FeO\underrightarrow{\left(2\right)}FeCl_2\underrightarrow{\left(3\right)}Fe\left(ỌH_2\right)\underrightarrow{\left(4\right)}FeSO_4\underrightarrow{\left(5\right)}FeCl_2\underrightarrow{\left(6\right)}Fe\left(NO_3\right)_2\)
Z/ \(Mg\left(OH\right)_2\underrightarrow{\left(1\right)}MgO\underrightarrow{\left(2\right)}MgSO_4\underrightarrow{\left(3\right)}MgCl_2\underrightarrow{\left(4\right)}Mg\left(OH\right)_2\underrightarrow{\left(5\right)}MgCl_2\underrightarrow{\left(6\right)}Mg\left(NO_3\right)_2\)
X/\(Al\left(OH\right)_3\underrightarrow{\left(1\right)}Al_2O_3\underrightarrow{\left(2\right)}AlCl_3\underrightarrow{\left(3\right)}Al\underrightarrow{\left(4\right)}Al_2\left(SO_4\right)_3\)
(1) Cu(OH)2 -t0-> CuO + H2O
(2) CuO + H2 -t0-> Cu + H2O
(3) Cu + Cl2 -t0-> CuCl2
(4) CuCl2 + 2NaOH - > Cu(OH)2 + 2NaCl
(5) Cu(OH)2 + H2SO4 - > CuSO4 + 2H2O
Cái nhóm (OH)2 thì xem như là 2 nguyên tử O và 2 nguyên tử H để cân bằng đúng ko ạ?