[pavel_sheva]

I am a new member so please help me ! 
Fe + F2   -->
P   + Br2  -->
Hg  + I2 -->
NH3 + Cl2 -->
H2S + Cl2 -->
[Ag(NH₃)₂]⁺ + Cl^- + H⁺ -->
Thanks for helping !

[agrobert]

You should read the forum rules and attempt these problems.

[Alpha-Omega]

OK I think you are a new Chemistry student....

Best way to approach this-predict products...is to determine the oxidation state (CHARGE) on your reactants....then try to stick them togeter in such a way that everything balances on the left and the right of the equation....

Like making an algebra eqution equal on both sides...

Try that first one....it is very easy....what can you make out of it if the oxidation state of Fe is +3/Fe+3 or if it is +2/Fe+2....try that approach

[Alpha-Omega]

OK so hint did not help or you are way lost.  Cannot leave this hanging so for your first RXN:

Fe + F2  ----> ?

Reaction of iron with the halogens:

Iron reacts with excess of the halogens F2, Cl2, and Br2, to form ferric, that is, Fe(III), halides.

2Fe(s) + 3F2(g) → 2FeF3(s) (white)

and a few more examples:

2Fe(s) + 3Cl2(g) → 2FeCl3(s) (dark brown)

2Fe(s) + 3Br2(l) → 2FeBr3(s) (reddish brown)

For educational purposes-This reaction is not very successful for iodine because of thermodynamic problems. The iron(III) is too oxidizing and the iodide is too reducing. The direct reaction between iron metal and iodine can be used to prepare iron (II) iodide, FeI2.

Fe(s) + I2(s) → FeI2(s) (grey)

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Reaction of phosphorus with the halogens:

White phosphorus, P4, reacts vigorously with all the halogens at room temeperature to form phosphorus trihalides. So, it reacts with fluorine, F2, chlorine, Cl2, bromine, Br2, and iodine, I2, to form respectively phosphorus(III) fluoride, PF3, phosphorus(III) chloride, PCl3, phosphorus(III) bromide, PBr3, and phosphorus(III) iodide, PI3.

[Alpha-Omega]

OOPS my mistake:

P + Br2 ----> ?

Reaction of phosphorus with the halogens:

White phosphorus, P4, reacts vigorously with all the halogens at room temeperature to form phosphorus trihalides. So, it reacts with fluorine, F2, chlorine, Cl2, bromine, Br2, and iodine, I2, to form respectively phosphorus(III) fluoride, PF3, phosphorus(III) chloride, PCl3, phosphorus(III) bromide, PBr3, and phosphorus(III) iodide, PI3.

P4(s) + 6Br2(g) → 4PBr3(l)

And just afew more:

P4(s) + 6F2(g) → 4PF3(g)

P4(s) + 6Cl2(g) → 4PCl3(l)

P4(s) + 6I2(g) → 4PI3(g)

And for educational purposes:

White phosphorus, P4, reacts with iodine, I2, in carbon disulphide (CS2) to form phosphorus(II) iodide, P2I4. The same compound is formed in the reaction between red phosphorus and iodine, I2, at 180°C.

P4(s) + 4I2(g) → 2P2I4(g)

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Hg + I2 ----> ?

Reaction of mercury with the halogens
Mercury metal reacts with fluorine, F2, chlorine, Cl2, bromine, Br2, or iodine, I2, to form the dihalides mercury(II) fluoride, HgF2, mercury(II) chloride, HgCl2, mercury(II) bromide, HgBr2, or mercury(II) iodide, HgI2, respectively.

Hg(l) + I2(s) → HgI2(s)

And just a few more:


Hg(l) + F2(g) → HgF2(s)

Hg(l) + Cl2(g) → HgCl2(s)

Hg(l) + Br2(l) → HgBr2(s)
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OK this is my I meet you half-way rule...so try the rest and Best of Luck...

[PanCerowany]

Answer to teh last one is pH dependent.