Ideal gas law equation.. please help (think I understand but maybe not)?

Molar mass of Mg(NO₃)₂ = (24.3 + 14.0×2 + 16.0×6) g/mol = 148.3 g/mol
Moles of Mg(NO₃)₂ = (3.74 × 10⁻² g) / (148.3 g/mol) = 2.522 × 10⁻⁴ mol

Equation for the reaction:
2Mg(NO₃)₂(g) → 2MgO(s) + 4NO₂(g) + O₂(g)
According to the equation, 2 moles of Mg(NO₃)₂ completely decomposes to give 5 moles of gas products (4 moles of NO₂ and 1 mole of O₂).
(Moles of Mg(NO₃)₂) : (Moles of gas produced) = 2 : 5
(2.522 × 10⁻⁴ mol) : (Moles of gas produced) = 2 : 5
Moles of the gas produced = (2.522 × 10⁻⁴ mol) × (5/2) = 6.305 × 10⁻⁴ mol

Consider the gas produced:
Moles, n = 6.305 × 10⁻⁴ mol
Pressure, P = 100 kPa
Temperature, T = (273.2 + 60.0) K = 333.2 K
Gas constant, R = 8.31 J K⁻¹ mol⁻¹ = 8.31 L kPa K⁻¹ mol⁻¹
Volume, V = ? L

Gas law: PV = nRT
Volume, V = nRT/P = (6.305 × 10⁻⁴) × 8.31 × 333.2 / 100 L = 0.0175 L = 17.5 mL

first of all.. via the ideal gas law.. PV = nRT ----> V = nRT/P
NOTE:.. volume is dependent on moles of gas present.. not the molar mass of the gas particles
.. .. .. .. .meaning.. ALL gas particles can be treated equally for this problem!

meaning
.. 2 Mg(NO3)2 ----> 5 moles of gas

*** part (a) ***
using that dimensional analysis you've recently mastered...

... 3.74x10^-2g Mg(NO3)2.. .. 1 mol Mg(NO3)2... . 5 mol gas
---- ---- ----- ---- ---- ----- ---- x ---- ---- ---- ----- ---- x ----- ----- ----- ----- = 6.305x10^-4 mol gas
... ... ... ... .. .. 1.. ... .... ... ... .. 148.3g Mg(NO3)2.. ... 2 mol Mg(NO3)2

then
.. PV = nRT
.. V = nRT/P
.. .. = (6.305x10^-4 mol) * (8.31 J/molK) * (08206 Latm / 8.314 J/mol) * (333.15 K) / (100 kPa * 1atm/101.325 kPa)
.. .. = 0.01746L * (1000 cm³ / L) = 17.5 cm³

You equation says for every 2 moles of Mg(NO3)2 you produce 4 moles of N0_2 (nitrous oxide) gas and 1 mole of O_2 gas

for every 2 moles of Mg(NO3)2 you get (4+1)= 5 moles of gas (combination of two different gases that is )
so 5 moles of gas / 2 moles of original compound