兩條簡單的化學選擇題

1.
The answer is B.

Ammonia solution is alkaline, because ammonia reacts with water to give hydroxide ions.
NH3(aq) + H2O(l) ≒ NH4+(aq) + OH-(aq)
The hydroxide ions react with iron(II) ions to give a green precipitate of iron(II) hydroxide.
Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)

The overall full equation can be written as:
2NH3(aq) + 2H2O(l) + FeSO4(aq) → (NH4)2SO4(aq) + Fe(OH)2(s)


A is incorrect, because there is no change in oxidation number.

B is correct, as mentioned above.

C is incorrect, because the ammonium and sulphate ions left in the solution is colourless. If ammonia solution is in excess, the solution is colourless. If iron(II) sulphate in excess, the solution is pale green.

D is incorrect, because there is no change in oxidation number, and the oxidation number of iron remains +2.


2.
The answer is B.

(1) is incorrect.
Both K2CO3 and Na2CO3 cannot be oxidized by potassium dichromate solution. However, the sulphuric acid that used to acidify the potassium permanate would reacts with all metal carbonates to give carbon dioxide gas. Since both potassium carbonate and sodium carbonate react to give the colourless gas of carbon dioxide, this method cannot be used to distinguish between potassium carbonate and sodium carbonate.

(2) is correct.
Sodium sulphite is a reducing agent, which can reduce orange acidified potassium dichromate solution to green chromium(III) ions.
3SO32-(aq) + Cr2O72-(aq) + 8H+(aq) → 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
Sodium sulphate is not a reducing agent, and thus it cannot react acidified potassium dichromate solution.

(3) is incorrect.
Both iron(II) sulphate and iron(II) nitrate contain iron(II) ions which reduce orange acidified potassium dichromate solution to green.
6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
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