# Reaction of Metallic Tin with NH4Cl and treatment with Sulfur

Metallic $\ce{Sn}$ is reacted with $\ce{NH_4Cl}$ and the resulting solution is treated with Sulfur.

1. What is the gaseous product?
2. What is the color of final precipitate?
3. The above resulting solution is added to $\ce{AuCl3}$, resulting in a purple coloured solution. What is the composition of the solution?

I do realise the answer to (3) will be the purple of Cassius i.e colloidal gold in $\ce{Sn(OH)2}$ (as per Vogel's Qualitative Inorganic Analysis textbook fifth edition). But I can't figure out the answer to the first two questions (I tried Googling with no fruitful search).

• Are you sure that the solution is treated with sulfur and not sulfide? – Loong Dec 26 '16 at 16:07
• Yes, I am not getting the purpose of treatment of sulfur. – Nilay Ghosh Dec 26 '16 at 16:17
• Yeah I know; me neither am sure of the purpose. But this what the question in my question paper said. – Utkarsh Gupta Dec 27 '16 at 7:37

Note: this is an incomplete answer because question is unclear till now

Metallic $\ce{Sn}$ is reacted with $\ce{NH4Cl}$..

$$\ce{2NH4Cl + Sn → SnCl2 (white ppt.) + 2NH3 (both aq and g.) + H2 ^}$$

Ammonium chloride react with tin to produce tin(II) chloride, ammonia and hydrogen. (source)

Some tin(II) chloride will react with aqueous ammonia to form tin (II) hydroxide and amm. chloride.(source)

$$\ce{SnCl2 + 2(NH3•H2O) or NH3(aq) → Sn(OH)2 or SnO.H2O(white ppt.) + 2NH4Cl (white ppt.)}$$

..and the resulting solution is treated with Sulfur.

I am suspecting that is hydrogen sulfide and not sulfur.(Are you sure?)

$$\ce{SnCl2 + H2S → SnS(dark brown ppt.) + HCl}$$

Tin(II) chloride react with hydrogen sulfide to produce tin(II) sulfide and hydrogen chloride. Tin(II) chloride - concentrated solution. Hydrogen sulfide - saturated solution. (source)

1. What is the gaseous product?

Hydrogen, Some ammonia gas.

1. What is the color of final precipitate?

Stannous sulfide ($\ce{SnS}$)- dark brown ppt.

1. The above resulting solution is added to $\ce{AuCl3}$, resulting in a purple coloured solution. What is the composition of the solution?

$$\ce{2[AuCl4]- + 4Sn^2+ + 2H2O -> [2Au + Sn(OH)2)](purple ppt.) + 3Sn^4+ + 2H+ + 8Cl-}$$

(Vogel)

Note: $\ce{AuCl3}$ exist as $\ce{AuCl4-}$ ions in solution.

• The answer says that the gaseous products are not just Hydrogen but Hydrogen and Ammonia – Utkarsh Gupta Dec 27 '16 at 7:39
• @UtkarshGupta Since ammonia is very soluble, some of it is present in aqueous form in the solution. Some may have passed as gas but most of the ammonia is in aqueous form. I will edit the answer though but are you sure it was sulfur and not hydrogen sulfide? – Nilay Ghosh Dec 27 '16 at 8:09
• I am pretty sure the question mentioned Sulfur (I have it in my hand right now). There is not much I can do about it. Thanks for your help :) – Utkarsh Gupta Dec 28 '16 at 12:11
• Ok, I will see what I can do.. – Nilay Ghosh Dec 28 '16 at 15:33