If 498 mol of octane combusts, what volume of carbon dioxide is produced at 23.0 °c and 0.995 atm?

Octane on combustion yields CO₂ and H₂O;

                         2C₈H₁₈  +  2 5O₂    →    16 CO₂  +  18 H₂O

According to equation,

                      2 moles Octane produces  =  16 moles of CO₂ 
So,
              498 moles Octane will produce  =  X moles of CO₂

Solving for X,
                                     X  =  (498 mol × 16 mol) ÷ 2 mol

                                     X  =  3984 moles of CO₂

Now, Calculating for Volume,
As,
                                    P V  =  n R T
Solving for V,
                                        V  =  n R T / P        (1)

Pressure  =  P  =  0.995 atm 
Volume  =  V  =  ? 
Moles  =  n  =  3984 
Gas Constant  =  R  =  0.0821 L.atm.K⁻¹.mol⁻¹ 
Temperature  =  T  =  23 + 273 = 296 K

Putting values in eq. 1

                V  =  (3984 mol × 0.0821 L.atm.mol⁻¹.K⁻¹ × 296 K) ÷ 0.995 atm

                                      Volume  =  973040.95 L

Only one reactant yields 2 or more products. meaning you will only have 1 product and it will break down into 2 or more products. eg. h2o -> h2 + o2