Answer:
2FeCl3(aq) + 3(NH4)2S(aq) →6NH4Cl(aq) + Fe2S3(s)
Explanation:
Step 1: Data given
The reactants are:
iron(III) chloride = FeCl3
ammonium sulfide = (NH4)2S
The products are:
ammonium chloride = NH4Cl
iron(III) sulfide = Fe2S3
Step 2: Balance the reaction
FeCl3(aq) + (NH4)2S(aq) → NH4Cl(aq) + Fe2S3(s)
On the right side we have 2x Fe, so on the left side, to balance the amount of Fe on both sides, we should multiply FeCl3 by 2
2FeCl3(aq) + (NH4)2S(aq) → NH4Cl(aq) + Fe2S3(s)
On the left side we have 6x Cl, on the right side 1x Cl. To balance the amount of Cl we have to multiply NH4Cl (on the right side) by 6
2FeCl3(aq) + (NH4)2S(aq) →6NH4Cl(aq) + Fe2S3(s)
On the right side we have 6x NH4, on the left side we have 2x NH4. To balance this amount we should multiply (NH4)2S on the left side by 3
Now the reaction is balanced.
2FeCl3(aq) + 3(NH4)2S(aq) →6NH4Cl(aq) + Fe2S3(s)
