Câu 1:Hoàn thành chuỗi phản ứng hóa học :

a) Na--->NaOH--->Na2CO3--->NaHCO3

--->NaOH--->NaCl---NaOH--->Na--->NaH--->

NaOH--->NaCl--->NaOCl.

b) Fe--->Fe2O3--->FeCl3--->Fe(OH)3--->

Fe2O3--->FeO--->FeSO4--->Fe.

c) S--->SO2--->H2SO4--->CuSO4--->K2SO3.

Đáp án + Giải thích các bước giải:

$a)$

$2Na+2H_2O\to 2NaOH+H_2\uparrow$

$CO_2+2NaOH\to Na_2CO_3+H_2O$

$Na_2CO_3+CO_2+H_2O\to 2NaHCO_3$

$NaHCO_3+Ba(OH)_2\to BaCO_3\downarrow+NaOH+H_2O$

$NaOH+HCl\to NaCl+H_2O$

$2NaCl+2H_2O\to 2NaOH+Cl_2\uparrow+H_2\uparrow$

$4NaOH\xrightarrow{đpnc}2H_2O+4Na+O_2\uparrow$

$2Na+H_2\xrightarrow{t^o}2NaH$

$2Na+2H_2O\to 2NaOH+H_2\uparrow$

$NaOH+HCl\to NaCl+H_2O$

$NaCl+H_2O\xrightarrow[kmn]{đpdd}H_2\uparrow+NaClO$

$b)$

$4Fe+3O_2\xrightarrow{t^o}2Fe_2O_3$

$Fe_2O_3+6HCl\to 2FeCl_3+3H_2O$

$FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl$

$2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O$

$Fe_2O_3+CO\xrightarrow{500^oC-600^oC}2FeO+CO_2\uparrow$

$FeO+H_2SO_4\to FeSO_4+H_2O$

$2Al+3FeSO_4\to Al_2(SO_4)_3+3Fe\downarrow$

$c)$

$S+O_2\xrightarrow{t^o}SO_2$

$2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3$

$SO_3+H_2O\to H_2SO_4$

$CuO+H_2SO_4\to CuSO_4+H_2O$

$a,\\ 2Na+2H_2O\to 2NaOH+H_2\\ 2NaOH+CO_2\to Na_2CO_3+H_2O\\ Na_2CO_3+CO_2+H_2O\to 2NaHCO_3\\ NaHCO_3+Ca(OH)_2\ dư\to CaCO_3+NaOH+H_2O\\ NaOH+HCl\to NaCl+H_2O\\ 2NaCl+2H_2O\xrightarrow[\text{màng ngăn xốp}]{đp dd} 2NaOH+Cl_2+H_2\\ 4NaOH\xrightarrow{đpnc} 4Na+O_2+2H_2O\\ 2Na+H_2\xrightarrow{t^0} 2NaH$

$NaH+H_2O\to NaOH+H_2\\ 2NaOH+CuCl_2\to Cu(OH)_2+2NaCl\\ NaCl+H_2O\xrightarrow[\text{không màng ngăn}]{đp dd}NaClO+H_2 $

$b,\\ 4Fe+3O_2\xrightarrow{t^0} 2Fe_2O_3\\ Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ FeCl_3+3NaOH\to Fe(OH)_3+3NaCl\\ 2Fe(OH)_3\xrightarrow{t^0} Fe_2O_3+3H_2O\\ Fe_2O_3+H_2\xrightarrow{t^0} 2FeO+H_2O\\ FeO+H_2SO_4\to FeSO_4+H_2O\\ FeSO_4+Mg\to MgSO_4+Fe$

$c,\\ S+O_2\xrightarrow{t^0} SO_2\\ SO_2+Cl_2+2H_2O\to H_2SO_4+2HCl\\ H_2SO_4+CuO\to CuSO_4+H_2O$