[Rutherford]

When 6.714g of an unknown metal M was dissolved in 25% HNO₃ 1500ml of a gas mixture was made. The mixture contains NO (39% of volume, 53.81% of mass), N₂O (21% of volume) and another gas A. Determine the metal M and the gas A.

I don't know where to start from first.

Edit: I made some mistakes, now I corrected the problem.

[Hunter2]

1. What reactions do you know, if nitric is used to dissolve a metals.
2. N₂O I dont beleave you can obtain, more suitable is NO₂
3. From the given volume you can calculate the mole of the gas. Ideal gas equation.

[Rutherford]

Okay, I calculated that the molar mass  of the gas is 2g/mol, meaning that it is H₂. How to determine the metal now?
Can't balance the equation because of H₂.

[Hunter2]

Which equation did you use.

example for copper http://en.wikipedia.org/wiki/Copper%28II%29_nitrate

The mole of the gas correspond to the moles of the metal.

[Rutherford]

M+HNO₃ MNO₃+N₂O+NO+H₂+H₂O
The problem is that it can't be balanced, probably because a mixture was made (the gasses weren't made in stochiometric amounts).

[Hunter2]

Nitrous oxide cannot optained by dissolving a metal in nitric acid.

http://en.wikipedia.org/wiki/Nitrous_oxide

The same with hydrogen. It will not survive in oxidising process. Only the solution is very diluted, but 25% is still half concentrated acid.

[Rutherford]

I don't know, that's the problem.

[Hunter2]

I got Indium as result. Hydrogen is correct.

In = 114 g/mol

You can calculate the mole of NO and N₂O. You can calculate the mass of both. With the percentatge of the mass its possible to get the Mass of A and also the Volume of A. The molare mass was correct 2 g/mol.

To put NO N₂O and H₂ in mole ratio you will get  4 HNO₃ to obtain the gasses. With this ratio you can figure out molare mass of the metal by given mass. The remaining oxygen correspond to M and by trying to get right factor the molare mass is optained.

[Rutherford]

H₂ is correct, Indium isn't. I think that you can't put NO N₂O and H₂ in mole ratio, because they make a mixture (mixtures have a random composition), so I run out of ideas.

[Borek]

To put NO N₂O and H₂ in mole ratio you will get  4 HNO₃ to obtain the gasses.

Please elaborate, it doesn't make much sense to me.

So far I don't see how it can be solved without unrealistic assumptions. But I have a feeling N₂O IS a product of reaction between nitric acid and some metal, but for the life of me I can't remember details.

[Hunter2]

ok

39% of 1500 ml is 585 ml

1 mol ~ 22,4 l  => 0,026 mol NO => 0,78 g,  MG = 30 g/mol

21% N₂O => 315 ml => 0,014 mol => 0,62 g, MG = 44 g/mol

o.78 g => 53.81%,  0.62 g => 42,78%  3,42 % left
3,42 % correspond to 0,049 g

100% -39% -21% = 40% 40% from 1500 ml = 600 ml

600 ml =>  0,026 mol  => MG = 1,9 ~ 2 g/mol  => H₂

ratio 0.026 NO , 0.014 N₂O and 0,026 H₂

means  1 NO , 0,5 N₂O and 1 H₂

or 2 NO , 1 N₂O and 2 H₂


this ratio correspond to 4 HNO₃

4 HNO₃ => 2 NO + N₂O + 2 H₂ + 9 O

this 9 O has to react with the M

2 NO => x M

0.78g /(2*30 g/mol) = 6.714 g/ x g/mol


M = 516.46 g/mol

The Oxide is type M₂O₃

The 9 O => 3 M₂O₃

We doubled the ratio above  9 means real 4,5


M = 516,46/ 4.5 = 114,7 g/mol and this fits to Indium what has an Oxide In₂O₃

Probably the oxide will be dissolved in the nitric to Indium nitrate.

[Borek]

4 HNO₃ => 2 NO + N₂O + 2 H₂ + 9 O

Hydrogen and oxygen being products of one decomposition reaction - no more questions from me.

[Rutherford]

It isn't an oxide that is dissolved, it is a metal. I thought that from a mixture, the number of moles of the metal that reacted cannot be determined, but it seems that it is so, because there is no other way. Your idea helped me to balance the equation (assuming that the metal is monovalent only the equivalent mass will be calculated):

18M+22HNO₃ 18MNO₃+2NO+N₂O+2H₂+9H₂O
Using stochiometry I got that the equivalent mass is 28.56g/mol, doesn't fit for any metal, if it was a divalent metal: equivalent mass*2=57.12g/mol nearest iron, but again nothing. Trivalent metal M=114.24g/mol which is realy near to In, so I thought that it should be the reasonable answer, but in fact the unknown metal is Mn! The divalent metal assumption was the nearest, but not completely accurate  .

[Borek]

It isn't an oxide that is dissolved, it is a metal.

That doesn't matter much - in both cases overall reaction would be a one producing water.

I thought that from a mixture, the number of moles of the metal that reacted cannot be determined, but it seems that it is so, because there is no other way.

As you wrote it yields a wrong answer, so you are just deluding yourself.

Your idea helped me to balance the equation (assuming that the metal is monovalent only the equivalent mass will be calculated):

18M+22HNO₃ 18MNO₃+2NO+N₂O+2H₂+9H₂O

Just because something can be balanced doesn't make it right.

Edit: I don't see a way to fit the amount of produced hydrogen if it is Mn. 6.714 g of Mn means 0.122 moles, there are 0.0268 moles of hydrogen. Way too much for my liking.

[AWK]

18M+22HNO3 = 18MNO3+2NO+N2O+2H2+9H2O

You should write down reactions of unknown metal oxidation separately - each reaction contains one product of reduction of nitric acid.
I do not believe that hydrogen can be a product in the presence of concentrated nitric acid. Moreover, the temperature and the pressure of gas mixture is missing.