Write the net ionic equation for the following reaction. Aqueous iron(III) sulfate is added to aqueous sodium sulfide to produce solid iron(III) sulfide and aqueous sodium sulfate.

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)

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i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)