Question

Consider three unlabeled bottles, each contain small pieces of one of the following metals. - Magnesium - Sodium - Silver The following reagents are used for identifying the metals. - Pure water - A solution of 1.0 molar HCl - A solution of concentrated HNO3 (a) Which metal can be easily identified because it is much softer than the other two? Describe a chemical test that distinguishes this metal from the other two, using only one of the reagents above. Write a balanced chemical equation for the reaction that occurs. (b) One of the other two metals reacts readily with the HCl solution. Identify the metal and write the balanced chemical equation for the reaction that occurs when this metal is added to the HCl solution. Use the table of standard reduction potentials (attached) to account for the fact that this metal reacts with HCl while the other does not. (c) The one remaining metal reacts with the concentrated HNO3 solution. Write a balanced chemical equation for the reaction that occurs. (d) The solution obtained in (c) is diluted and a few drops of 1 M HCl is added. Describe what would be observed. Write a balanced chemical equation for the reaction that occurs.

Answer 1

a) Sodium has one valence electron so it does not form a strong metallic bond. Sodium has a large atomic size. Therefore, because of its one valance electron and its relatively larger size, Na is a soft metal. Na is also highly reactive and it is never found naturally on earth.

Sodium reacts vigorously with water to form sodium hydroxide and hydrogen gas.

2Na + 2H₂O → 2NaOH + H₂

b) As per the activity series of metals:

K > Na > Ca > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Hg > Ag > Au

Mg is more reactive than Ag. Ag is a noble metal and does not react with 1M HCl

Mg reacts with HCl to form magnesium chloride and water.

Mg + 2HCl → MgCl₂ + H₂

c) Silver reacts with concentrated nitric acid to form silver nitrate, nitrogen dioxide and water.

Ag + 2HNO₃ → AgNO₃ + NO₂ + H₂O

d) When 1M HCl is added to solution c. One of the products of solution c is AgNO₃. So AgNO₃ will react with HCl to form a white precipitate of AgCl and nitric acid.

AgNO₃ + HCl → AgCl + HNO₃

Answer 2

Step-by-step explanation:

Reagents: Water , 1.0 M HCl, Conc.
`HNO_3`

Metal-1 = Magnesium

Metal-2 = Sodium

Metal-3 = Silver

a) Out of these three metals sodium is a soft metal which can be easily cut with help of knife.Sodium metal reacts violently with water to form hydrogen gas and sodium hydroxide.

`Na(s)+H_2O(l)\rightarrow NaOH(aq)+H_2(g)`

b) Out of magnesium and silver, magnesium metal reacts readily with HClto form magnesium chloride and hydrogen gas. Where as silver being noble metal doesn't get attacked by the HCl.

`Mg(s)+2HCl(aq)\rightarrow MgCl_2+H_2`

c) Silver is a noble metal. And noble metals only react with concentrated
`HNO_3` that is nitric acid.

`3Ag(s)+HNO_3(aq)\rightarrow 3AgNO_3(aq)+2H_2O(l)+NO(g)`

d) When few drops of HCl were added to the solution of silver and
`HNO_3`. This will leads to the formation of nascent chloride which reacts with silver metal present in the solution to give silver chloride.

`1HNO_3+3HCL\rightarrow NOCl+2H_2O+2Cl(\text{nascent chloride})`

`Ag+Cl\rightarrow AgCl`