Respuesta :
[tex]Fe_{3}So_{4} + Na_{2}S = Fe_{3}S + Na_{2}SO_{4}[/tex]
Explanation:
Formula for reactants and products given is:
- Aqueous sodium sulfide = [tex]Na_{2}S[/tex]
- Iron(III) sulfide = [tex]Fe_{2}S[/tex]
- Aqueous sodium sulfate = [tex]Na_{2}SO_{4}[/tex](aq)
- Aqueous iron(III) sulphate = [tex]Fe_{3}SO_{4}[/tex](aq)
Therefore;
The net ionic equation for the given reaction is as follows:
[tex]Fe_{3}So_{4} + Na_{2}S = Fe_{3}S + Na_{2}SO_{4}[/tex]
Answer:
2Fe^3+(aq) + 3S^2-(aq) → Fe2S3(s)
Explanation:
Step 1: Data given
iron(III) sulfate = Fe2(SO4)3
sodium sulfide = Na2S
iron(III) sulfide = Fe2S3
sodium sulfate = Na2SO4
Step 2: The unbalanced equation
Fe2(SO4)3 + Na2S → Fe2S3 + Na2SO4
Step 3: Balancing the equation
Fe2(SO4)3 + Na2S → Fe2S3 + Na2SO4
On the left side we have 12x O ( in Fe2(SO4)3), on the right side we have 4x O (in Na2SO4). To balance the amount of O on both sides, we have to multiply Na2SO4 (on the right) by 3.
Fe2(SO4)3 + Na2S → Fe2S3 + 3Na2SO4
Now, we have 6x Na on the right side, and 2x Na on the left side. To balance the amount of Na on both side, we have to multiply Na2S by 3.
Now the equation is balanced.
Fe2(SO4)3(aq) + 3Na2S(aq) → Fe2S3(s) + 3Na2SO4(aq)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will look like this
2Fe^3+(aq) + 3S^2-(aq) → Fe2S3(s)
