To solve the question we will assume that the gas behaves like an ideal gas, that is to say, that there is no interaction between the molecules. Assuming ideal gas we can apply the following equation:

Where,
P is the pressure of the gas
V is the volume of the gas
n is the number of moles
R is a constant
T is the temperature
Now, we have two states, an initial state, and a final state. The conditions for each state will be.
Initial state (1)
P1=975Torr=1.28atm
V1=3.8L
T1=-18°C=255.15K
Final state(2), STP conditions
P2=1atm
T2=273.15K
V2=?
We will assume that the number of moles remains constant, so the nR term of the first equation will be constant. For each state, we will have:

Since nR is the same for both states, we can equate the equations and solve for V2:

We replace the known values:

At STP conditions the gas would occupy 5.2L. First option
When 2 moles burn 6 moles of O2 is produced !
so for 16 moles , (6/2)×16= 48 moles will be produced !
so answer is A , 48 moles !
Since the reaction has one mole of reactants versus two moles of products, the reaction would have a decrease in entropy. One mole of gas exerts less pressure than two moles of gas, and therefore one mole of gas has more entropy than two moles of gas.
Hope this helps