I have some queries regarding the following:
2FeCl3 (aq) + H2S (g) -> 2FeCl2 (aq) + 2HCl (aq) + S (s)
Its ionic equation can also be written as:
2Fe 3+ + 6Cl- + H2S -> 2Fe 2+ + 4Cl- + 2H+ + 2Cl- + S
Looking at the first reactant. FeCl3... since chlorine has 3 atoms, and has a "2" in front, do I multiply 3 to 2 to get 6Cl- then?
In this case, can I cancel out 6Cl-, 4Cl- and 2Cl-?
Lots of thanks.
If I remember correctly (shit need to practise ionic equations liao),
Chem Eq: 2FeCl3 (aq) + H2S (g) -> 2FeCl2 (aq) + 2HCl (aq) + S (s)
Ionic Eq: 2Fe 3+ + 6Cl - + H2S -> 2Fe 2+ + 4Cl- + 2H+ + 2Cl- + S
2Fe 3+ + H2S -> 2Fe 2+ + 2H+ + S
In other words, I think you're right.