What is the mass in the original mixture?

On strong heating :
2NaNO₃(s) → 2NaNO₂(s) + O₂(g)
2KNO₃(s) → 2KNO₂(s) + O₂(g)

Mole ratio NaNO₃ : O₂ = 2 : 1
Mole ratio KNO₃ : O₂ = 2 : 1

As Mole ratio NaNO₃ : KNO₃ = 1 : 2
Then Mole ratio (O₂ formed from NaNO₃) : (O₂ formed from KNO₃) = 1 : 2

Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
Total no. of moles of O₂ formed = (19.2 g) / (32.0 g/mol) = 0.600 mol

No. of moles of O₂ formed from NaNO₃ = (0.600 mol) × [1/(1+2)] = 0.200 mol
Molar mass of NaNO₃ = (23.0 + 14.0 + 16.0×3) g/mol = 85.0 g/mol
Mass of NaNO₃ in the mixture = (85.0 g/mol) × (0.200 mol) = 17.0 g

No. of moles of O₂ formed from KNO₃ = (0.600 - 0.200) mol = 0.400 mol
Molar mass of KNO₃ = (39.0 + 14.0 + 16.0×3) g/mol = 101.0 g/mol
Mass of KNO₃ in the mixture = (101.0 g/mol) × (0.400 mol) = 40.4 g

Total mass of the mixture = (17.0 + 40.4) g = 57.4 g

Supposing the reactions to be something like:
4 NaNO3 → 2 Na2O + 2 N2 + 5 O2
4 KNO3 → 2 K2O + 2 N2 + 5 O2

For ease of calculation, rewrite the same reactions as:
4 NaNO3 → 2 Na2O + 2 N2O5
4 KNO3 → 2 K2O + 2 N2O5

(19.2 g N2O5) / (108.0104 g N2O5/mol) = 0.17776 mol N2O5

Since the mole ratio of NaNO3 to KNO3 is 1 to 2, then the mole ratio of N2O5 from NaNO3 to N2O5 from KNO3 is also 1 to 2:

(1) / (1 + 2) x (0.17776 mol N2O5 total) = 0.059253 mol N2O5 from NaNO3
(2) / (1 + 2) x (0.17776 mol N2O5 total) = 0.11851 mol N2O5 from KNO3

(0.059253 mol N2O5) x (4 mol NaNO3 / 2 mol N2O5) x (84.9947 g NaNO3/mol) = 10.072 g NaNO3
(0.11851 mol N2O5) x (4 mol KNO3 / 2 mol N2O5) x (101.1032 g KNO3/mol) = 23.963 g KNO3

10.072 g NaNO3 + 23.963 g KNO3 = 34.0 g total in the original mixture

[I have the feeling that there's an easier way to go about this problem, but I'm not seeing it just now.]

Yeah, it depends on the decomposition.........

Two different answers, based on different decomposition reactions. Which to choose?

I found this response to a related question to be interesting:

http://www.madsci.org/posts/archives/1999-09/937717553.Ch.r.html