Many metal ions form insoluble sulfide compounds when a solution of the metal ion is treated with hydrogen sulfide gas. For example, nickel(II) precipitates nearly quantitatively as NiS when H2S gas is bubbled through a nickel ion solution. How many mL of gaseous H2S at STP are needed to precipitate all the nickel ion present in 10mL of 0.050M NiCl2 solution?
2006-12-04 02:38:07 · 2 個解答 · 發問者 Anonymous in 科學 ➔ 化學
NiCl2 + H2S → NiS + 2HCl
Mole ratio NiCl2 : H2S = 1 : 1
No. of moles of NiCl2 used = 0.050 x (10/1000) = 0.0005 mol
No. of moles of H2S needed = 0.0005 mol
Molar volume of H2S at stp = 22400 mL/mol
Volume of H2S needed at stp = (0.0005) (22400) = 11.2 mL
2006-12-04 06:32:32 · answer #1 · answered by Uncle Michael 7 · 0⤊ 0⤋
You should be able to find the data from "CRC Handbook of Chemistry and Physics" from library or from the appendix in your textbook, or data given by your teacher, etc. Since I don't have a CRC Handbook nor a textbook that has this data, I can't give you the data you need.
2006-12-04 06:03:52 · answer #2 · answered by CHL 2 · 0⤊ 0⤋