Question

Question #bd074

Answer 1

See below:

Explanation:

`sf(8(a). (i))`

Potassium iodide solution could be identified by its reaction with iron(III) sulfate solution.

This is a redox reaction where `sf(I^-)` is oxidised to `sf(I_2)` by the `sf(Fe^(3+))` ions.

Omitting the spectator ions:

`sf(Fe_((aq))^(3+)+I_((aq))^(-)rarrFe_((aq))^(2+)+(1)/2I_(2(aq)))`

Addition of starch solution would give a blue/black precipitate to confirm the presence of iodine since iron(III) and iodine are both brown in colour.

`sf(---------------------)`

Barium chloride solution would give a white precipitate with iron(III) sulfate solution of barium sulfate. Again, omitting the spectators:

`sf(Ba_((aq))^(2+)+SO_(4(aq))^(2-)rarrBaSO_(4(s)))`

Strictly speaking you should add an XS of dilute HCl to dissolve other possible precipitates.

`sf(---------------------)`

If you mix iron(III) sulfate solution with the `sf(K_4Fe(CN)_6)` solution you get a dark blue solid commonly known as "Prussian Blue":

`sf(Fe^(3+)+[Fe^(II)(CN)_6]^(4-)rarr[Fe^(III)[Fe^(II)(CN)_6]]^(-))`

`sf(---------------------)`

`sf((a)(ii))`

Aluminium will dissolve in warm sodium hydroxide solution to give sodium aluminate and hydrogen:

`sf(2Al_((s))+2NaOH_((aq))+2H_2O_((l))rarr2NaAlO_(2(aq))+3H_(2(g)))`

This reflects the amphoteric nature of aluminium.

The same type of reaction occurs with zinc which is also amphoteric:

`sf(Zn(s)+H_2O(l)+2NaOH(aq)rarrNa_2Zn(OH)_2(aq)+H_2(g))`

If ammonium ions are warmed with hydroxide ions, ammonia gas is evolved which will turn red litmus blue:

`sf(NH_(4(aq))^++OH_((aq))^(-)rarrNH_(3(g))+H_2O_((l)))`

An aqueous solution of chlorine will react with sodium hydroxide solution, removing any green colouration due to elemental chlorine:

`sf(Cl_(2(aq))+2NaOH_((aq))rarrNaCl_((aq))+NaOCl_((aq))+H_2O_((l)))`

This is an example of "disproportionation". Chlorine (0) is simultaneously oxidised to `sf(OCl^-)` (+1) and to `sf(Cl^-)` (-1).

`sf((b)(i)`

`sf(A)` is ammonium chromate(VI) and decomposes on heating to give green chromium(III) oxide:

It is quite spectacular and is known as "The Volcano Experiment".

`sf((NH_4)_2Cr_2O_(7(s))rarrCr_2O_(3(s))+N_(2(g))+4H_2O_((g))`

So `sf(M)` is chromium.

`sf(B)` is chromium(III) oxide `sf(Cr_2O_3)`.

`sf(C)` is nitrogen gas `sf(N_2)`.

`sf(N_2)` will burn in magnesium to give magnesium nitride:

`sf(3Mg_((s))+N_(2(g))rarrMg_3N_(2(s))`

So `sf(D)` is magnesium nitride.

This reacts with water:

`sf(Mg_3N_(2(s))+6H_2O_((l))rarrMg(OH)_(2(s))+2NH_3(g))`

So `sf(E)` is ammonia, which turns red litmus paper blue.

When aqueous `sf(A)` is warmed with carbonate ions an acid base reaction occurs giving `sf(E)` which we know is ammonia:

`sf(NH_(4(aq))^++CO_(3(aq))^(2-)rarrNH_(3(g))+HCO_(3(aq))^-)`

Green chromium(III) oxide is also amphoteric and will dissolve in aqueous alkali to give `sf([Cr(OH)_6]^(3-))` which is soluble.

Under these conditions hydrogen peroxide is a powerful oxidising agent and is able to oxidise green `sf(Cr(III))` to give yellow `sf(Cr(VI))`:

`sf(2[Cr(OH)_6]^(3-)+3H_2O_2rarr2CrO_4^(2-)+2OH^(-)+8H_2O)`