[Borek]

You should write down reactions of unknown metal oxidation separately - each reaction contains one product of reduction of nitric acid.

Interesting idea - but looks like it leads nowhere. Assuming Mn²⁺ is being produced each time, and the gas mixture (I am assuming STP) contains 0.026 moles of NO₂, 0.014 moles of N₂O and 0.027 moles of H₂, total mass of manganese dissolved is 1.48+3.08+0.714=5.27g, less than 6.714g given in the question.

But as you have already noted, hydrogen is a very unlikely product in these conditions, as it reduces the nitric acid. That means products are rather unpredictable.

[Hunter2]

N₂O as well is unlikely. NO₂ can be formed.

[AWK]

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.

2 HNO3 + 8 HCl + 4 SnCl2 → 5 H2O + 4 SnCl4 + N2O
This is the reaction from wikipedia link given by Hunter2

[AWK]

I am using Na as example, but in analogy this eventually leads to equivalents of unknown metal.
3Na + 4HNO₃ = 3NaNO₃ + NO + 2H₂O
8Na + 10HNO₃ = 8NaNO₃ + N₂O + 5H₂O
10Na + 12HNO₃ = 10NaNO₃ + N₂ + 6H₂O
Only these 3 reactions above should be taken into account since these gases are poorly soluble in water.
Traces (but only traces) of hydrogen can be detected in gaseous mixture only only if acid was very diluted and reaction not mixed. But problem is still unsolvable!
4Na + 6HNO₃ = 4NaNO₃ + N₂O₃ + 3H₂O
Na + 2HNO₃ = NaNO₃ + NO₂ + H₂O
N₂O₃ and NO₂ are soluble in water (25 % HNO₃ contains it), moreover water is produced in each reaction - both react with water).
8Na + 10HNO₃ = 8NaNO₃ + NH₄NO₃ + 3H₂O
This reaction can be expected only in diluted acid.

[Borek]

2 HNO3 + 8 HCl + 4 SnCl2 → 5 H2O + 4 SnCl4 + N2O
This is the reaction from wikipedia link given by Hunter2

Technically it is not a metal dissolution, but reduction with tin(II) chloride.

[Borek]

3Na + 4HNO₃ = 3NaNO₃ + NO + 2H₂O
8Na + 10HNO₃ = 8NaNO₃ + N₂O + 5H₂O
10Na + 12HNO₃ = 10NaNO₃ + N₂ + 6H₂O

As I wrote earlier - this approach (with Mn²⁺ being a product) shows only 5.27g of the Mn is dissolved, so it doesn't fit the answer given.

I don't get your last equation - while technically it is balanced, we don't have nitrogen between products. Did you mean H₂?

[AWK]

I don't get your last equation - while technically it is balanced, we don't have nitrogen between products. Did you mean H2?

As I wrote previously hydrogen in the presence of medium concentrated nitric acid is imposible. Hydrogen is an idea of Raderford - not information from data.

[Borek]

As I wrote previously hydrogen in the presence of medium concentrated nitric acid is imposible. Hydrogen is an idea of Raderford - not information from data.

Yes, it is impossible, no, it is not Raderford's random idea - it IS from the data. Assuming STP and given composition of the gas mixture, molar mass of the gas A can be calculated to be 2. It makes it hydrogen, doesn't it?

[Rutherford]

Hydrogen is the gas A in the answer.
I don't have any more idea here. If someone by some logic way calculates that the metal should be Mn, it would be nice to show it.

[AWK]

36 years I am  fighting with students that they write reactions with nitric acid without hydrogen or hydrogen sulfide as a product of reaction. Unfortunately some textbooks exist with such horrible errors and my hard  work is still useless.
The next problem is rounding intermediate results to one or two significant digits (as Hunter2 did). This also may cause  horrible errors. Rounding is the last mathematical operation in our calculations!
Of course in this problem from these data  the molsr mass  of unknown gas is 2.(but in real chemistry this cannot be hydrogen).

3Na + 4 HNO3 = 3NaNO3 + NO + 2H2O
8Na + 10 HNO3 = 8NaNO3 + N2O + 5H2O
2Na + 2HNO3 = 2NaNO3 + H2
After identification numer of moles of all gases we can calculate number of equivalents of monovalent “element” (3 x moles of NO + 8 x moles of N2O + 2 x moles of H2). Dividing mass of metal by number of equivalents we will get mass of one equivalent. Equivalent multiplied by 1, 2, 3 … should give  the mass of real element.

[Rutherford]

This gives the nearest answer to Mn (I got 55.49g/mole, near to Fe, too). Probably this problem was meant to be solved this way, so thank you AWK.

[Borek]

0.026 moles of NO₂

Sigh, should be NO. That's why I was getting incorrect results.