Question #bd074

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Question #bd074

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Answer 1 · 2016-07-07T22:40:09

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Explanation:

$8 (a) . (i)$ $sf(8(a). (i))$

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

This is a redox reaction where $I^{-}$ $sf(I^-)$ is oxidised to $I_{2}$ $sf(I_2)$ by the $F e^{3 +}$ $sf(Fe^(3+))$ ions.

Omitting the spectator ions:

$F e_{(a q)}^{3 +} + I_{(a q)}^{-} \to F e_{(a q)}^{2 +} + \frac{1}{2} I_{2 (a q)}$ $sf(Fe_((aq))^(3+)+I_((aq))^(-)rarrFe_((aq))^(2+)+(1)/2I_(2(aq)))$

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:

$B a_{(a q)}^{2 +} + S O_{4 (a q)}^{2 -} \to B a S O_{4 (s)}$ $sf(Ba_((aq))^(2+)+SO_(4(aq))^(2-)rarrBaSO_(4(s)))$

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 $K_{4} F e {(C N)}_{6}$ $sf(K_4Fe(CN)_6)$ solution you get a dark blue solid commonly known as "Prussian Blue":

$F e^{3 +} + {[F e^{I I} {(C N)}_{6}]}_{6}^{4 -} \to {[F e^{I I I} [F e^{I I} {(C N)}_{6}]]}^{-}$ $sf(Fe^(3+)+[Fe^(II)(CN)_6]^(4-)rarr[Fe^(III)[Fe^(II)(CN)_6]]^(-))$

$- - - - - - - - - - - - - - - - - - - - -$ $sf(---------------------)$

$(a) (i i)$ $sf((a)(ii))$

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

$2 A l_{(s)} + 2 N a O H_{(a q)} + 2 H_{2} O_{(l)} \to 2 N a A l O_{2 (a q)} + 3 H_{2 (g)}$ $sf(2Al_((s))+2NaOH_((aq))+2H_2O_((l))rarr2NaAlO_(2(aq))+3H_(2(g)))$

This reflects the amphoteric nature of aluminium.

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

$Z n (s) + H_{2} O (l) + 2 N a O H (a q) \to N a_{2} Z n {(O H)}_{2} (a q) + H_{2} (g)$ $sf(Zn(s)+H_2O(l)+2NaOH(aq)rarrNa_2Zn(OH)_2(aq)+H_2(g))$

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

$N H_{4 (a q)}^{+} + O H_{(a q)}^{-} \to N H_{3 (g)} + H_{2} O_{(l)}$ $sf(NH_(4(aq))^++OH_((aq))^(-)rarrNH_(3(g))+H_2O_((l)))$

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

$C l_{2 (a q)} + 2 N a O H_{(a q)} \to N a C l_{(a q)} + N a O C l_{(a q)} + H_{2} O_{(l)}$ $sf(Cl_(2(aq))+2NaOH_((aq))rarrNaCl_((aq))+NaOCl_((aq))+H_2O_((l)))$

This is an example of "disproportionation". Chlorine (0) is simultaneously oxidised to $O C l^{-}$ $sf(OCl^-)$ (+1) and to $C l^{-}$ $sf(Cl^-)$ (-1).

$(b) (i)$ $sf((b)(i)$

$A$ $sf(A)$ 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".

${(N H_{4})}_{2} C r_{2} O_{7 (s)} \to C r_{2} O_{3 (s)} + N_{2 (g)} + 4 H_{2} O_{(g)}$ $sf((NH_4)_2Cr_2O_(7(s))rarrCr_2O_(3(s))+N_(2(g))+4H_2O_((g))$

So $M$ $sf(M)$ is chromium.

$B$ $sf(B)$ is chromium(III) oxide $C r_{2} O_{3}$ $sf(Cr_2O_3)$ .

$C$ $sf(C)$ is nitrogen gas $N_{2}$ $sf(N_2)$ .

$N_{2}$ $sf(N_2)$ will burn in magnesium to give magnesium nitride:

$3 M g_{(s)} + N_{2 (g)} \to M g_{3} N_{2 (s)}$ $sf(3Mg_((s))+N_(2(g))rarrMg_3N_(2(s))$

So $D$ $sf(D)$ is magnesium nitride.

This reacts with water:

$M g_{3} N_{2 (s)} + 6 H_{2} O_{(l)} \to M {g (O H)}_{2 (s)} + 2 N H_{3} (g)$ $sf(Mg_3N_(2(s))+6H_2O_((l))rarrMg(OH)_(2(s))+2NH_3(g))$

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

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

$N H_{4 (a q)}^{+} + C O_{3 (a q)}^{2 -} \to N H_{3 (g)} + H C O_{3 (a q)}^{-}$ $sf(NH_(4(aq))^++CO_(3(aq))^(2-)rarrNH_(3(g))+HCO_(3(aq))^-)$

Green chromium(III) oxide is also amphoteric and will dissolve in aqueous alkali to give ${[C r {(O H)}_{6}]}_{6}^{3 -}$ $sf([Cr(OH)_6]^(3-))$ which is soluble.

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

$2 {[C r {(O H)}_{6}]}_{6}^{3 -} + 3 H_{2} O_{2} \to 2 C r O_{4}^{2 -} + 2 O H^{-} + 8 H_{2} O$ $sf(2[Cr(OH)_6]^(3-)+3H_2O_2rarr2CrO_4^(2-)+2OH^(-)+8H_2O)$

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Question #bd074

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