Small pieces of silver coloured metal X were added to concentrated HNO3. A brown has Z and a colourless solution containing salt Y were formed. Analysis of a 0.0914 mol sample of Z showed 1.28 grams of nitrogen gas and 2.93 grams of oxygen gas. The small sample of the colourless solution was diluted in water and divided into two portions. In one portion, aqueous NaOH was added drop by drop until it was in excess. A white precipitate W was formed that redissolved in excess NaOH. To another portion, aqueous NH3 was added drop by drop until it was in excess. A white precipitate P was formed that redissolved in excess NH3.
a)i) Name the white precipitate W.
ii) Construct the ionic equation with state symbol for the formation of W.
b) Name X and Y.
c) Calculate the relative atomic mass for gas Z.
d) How the analytical test of nitrate ion differ from ammonium ion?
Answers
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Answer:
i) Zinc Hydroxide Zn(OH)2
ii) Zn^+2 + 2OH^- = Zn(OH)2
b) X is Zinc
Y is Zinc Nitrate
c) Mr = 46
d) analytical test for nitrate ion is to add aqeous sodium hydroxide, then add aluminium and warm the mixture carefully. A gas is produced. (ammonia gas). This test is for anion.
The test for ammonia gas is to test it with litmus paper. It turns damp red litmus paper blue. while this test is for gas
Explanation:
i) metal (zinc) reacts with (sodium) hydroxide to give of metal hydroxide (zinc hydroxide)
b) zinc metal is the only metal cation which gives white ppt and is soluble in excess when added to sodium hydroxide and aqueous ammonia.
c) molecular formula of the gas = NO2
N : O
1.28 g : 2.93g
= 0.091429 mol : 0.183125 mol
= 1 : 2 (1 mole of nitrogen and 2 moles of oxygen).
Mr = Ar (nitrogen) + Ar (oxygen)
= 14 + 16 +16
= 46.