CobaltChloride - 29-4-2018 at 09:50
I dissolved some tin metal in hydrochloric acid contaminated with iron III chloride and left the solution outside. Initially, the hydrochloric acid
turned colorless from the redox between Fe3+ and Sn2+ that makes the much less intensely colored Fe2+ and Sn4+. Afterwards I left it to evaporate and
I was left with a white powder with a slight brown tinge, meaning that all Fe2+ (and subsequently Sn2+) was oxidized. Now I have one question: What is
the powder? Is it SnCl4*5H2O or SnO2 made by the hydrolysis of the former?
[Edited on 29-4-2018 by CobaltChloride]
woelen - 30-4-2018 at 11:59
SnCl4 almost certainly will hydrolyze to SnO2 when evaporated to dryness.
But to be sure, try adding the powder to some dilute HCl. If it is SnCl4.5H2O, then it will dissolve and you get a clear liquid, if it is SnO2 you get
a cloudy liquid.
CobaltChloride - 4-5-2018 at 09:21
Sorry I didn't answer you earlier. It seems that there are also some colorless needle-like crystals besides the very slightly yellowish plate-like
ones. Both of them dissolve in 10% HCl and give a colorless solution. The solution of needle crystals reduces Fe III to Fe II, so that should be
SnCl2.2H2O. The solution of plate crystals doesn't do that, so it should be SnCl4.5H2O.