A gas mixture is made by combining 6.8 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occu?

pv=nrt -> solve for V with STP, n=1. -> ~ 22.4L, or 0.0224m^3
thus 1 mol of an ideal gas occupies about 22.4L.

figure out how many mols of argon and neon you have (each 6.8g, argon is about 40g/mol and neon about 20g/mol)
16.12L/22.4L is the # of mol you can put in the volume. thus you can subtract the #mol of argon and neon that you know and come up with the mole fraction of your last, unknown, gas.
then you multiplly the mole fraction of your unknown gas with 6.8g and come up with the amu for the diatomic molecule. divide by 2 (since it's diatomic) to arrive at the molecular weight of one atom of your unknown gas.

The gas mixture :
P = 1 atm
V = 16.12 L
n = ? mol
R = 0.0821 atm L / (mol K)
T = 273 K

PV = nRT
n = PV/(RT) = 1 × 16.12 / (0.0821 × 273) = 0.7192 mol

Molar mass of Ar = 39.95 g/mol
Molar mass of Ne = 20.18 g/mol
Let molar mass of the unknown diatomic gas = y g/mol

Total number of moles :
(6.8/y) + (6.8/39.95) + (6.8/20.18) = 0.7192
1/y = (0.7198/6.8) - (1/39.95) - (1/20.18)
y = 32.1

Molar mass of the unknown diatomic gas = 32.1 g/mol

.. ... ... .. .. .. .16.12L.. .. ...1 mol
moles mix = ---- ----- x ---- ----- ---- --- = 0.71932
... ... .... ... ... .. ..1.. .. ..22.41L@STP

moles Ar = 6.8g x (1 mol / 39.9g) = 0.170

moles Ne = 6.8g x (1 mol / 20.8g) = 0.337

moles unknown = 0.71932 - 0.170 - 0.337 = 0.212 moles

molar mass of diatomic molecule = (6.8g / 0.212mol) = 32 g/mol... (2 sig figs)

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now.. you might be tempted to jump to the conclusion that your gas is O2.. but keep in mind there are 2 kinds of diatomic gases
.. (1) homonuclear... both atoms are of the same element
.. (2) heteronuclear... the two atoms are of different elements

CO is a diatomic gas
O2 is a diatomic gas
HCl is a diatomic gas...
etc

you have aids