Đáp án:

`↓`

Giải thích các bước giải:

`-` `MnO_2``+``4HCl``→``MnCl_2``+``Cl_2↑``+``2H_2O`

`-` `BaCl_2` $\xrightarrow[]{đpnc}$ `Ba``+``Cl_2↑`

`-` `H_2``+``Cl_2` $\xrightarrow[]{as}$ `2HCl`

`-` `Fe``+``2HCl``→``FeCl_2``+``H_2↑`

`-` `FeCl_2``+``2NaOH``→``Fe(OH)_2↓``+``2NaCl`

`-` `2Fe``+``3Cl_2` $\xrightarrow[]{t^o}$ `2FeCl_3`

`-` `FeCl_3``+``3NaOH``→``Fe(OH)_3↓``+``3NaCl`

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

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`-` `C``+``CO_2` $\rightleftharpoons$ `2CO↑`

`-` `2CO``+``O_2` $\xrightarrow[]{t^o}$ `2CO_2↑`

`-` `CO_2``+``NaOH``→``NaHCO_3`

`-` `2NaHCO_3` $\xrightarrow[]{t^o}$ `Na_2CO_3``+``CO_2↑``+``H_2O`

`-` `Na_2CO_3``+``2HCl``→``2NaCl``+``CO_2↑``+``H_2O`

`-` `CO_2``+``Ba(OH)_2``→``BaCO_3↓``+``H_2O`

`-` `BaCO_3``+``2HCl``→``BaCl_2``+``CO_2↑``+``H_2O`