Write balance molecular and net ionic equation `Fe_2(CO_3)_3 (s)+ HCl + Na_2S (aq)` yields...

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llltkl | College Teacher | (Level 3) Valedictorian

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In this reaction, a black precipitate of `Fe_2S_3` is formed. The overall reaction occurs in three steps:

1. Acid-base reaction: `Fe_2(CO_3)_3+6HCl rarr 2FeCl_3 + 3H_2CO_3`

2. Decomposition reaction: `3H_2CO_3 rarr 3CO_2 (g) + 3H_2O`

3. Precipitation reaction: `2FeCl_3+3Na_2S rarr Fe_2S_3(s)+6NaCl`


Overall: `Fe_2(CO_3)_3+6HCl+ 3Na_2S rarr Fe_2S_3(s)+6NaCl + 3CO_2 (g) + 3H_2O`

The equation can be written in the ionic form as:

`2Fe^(3+)(aq)+3CO_3^(2-)(aq)+6H^(+)(aq)+6Cl^(-)(aq)+ 6Na^(+)(aq)+3S^(2-)(aq) rarr Fe_2S_3(s)+6Na^(+)(aq)+6Cl^(-)(aq) + 3CO_2 (g) + 3H_2O`  

Cancelling similar terms from both sides, the net ionic equation can thus be written as:

`2Fe^(3+)(aq)+3CO_3^(2-)(aq)+6H^(+)(aq)+3S^(2-)(aq) rarr Fe_2S_3(s)+ 3CO_2 (g) + 3H_2O`


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