If 465mL of oxygen at STP is used in the reaction, what volume of CO2, measured at 37.3 degrees Celsius and 0.973atm, is produced?

It has been found that the reaction involved is :
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

At STP, 5 volume of O₂ produces 3 volumes of CO₂.
At STP, volume of CO₂ produced = 465 × (3/5) = 279 mL


Method 1 :

1 mole of gas occupies 22400 mL of volume.
No. of moles of O₂ used = 465/22400 = 0.02076 mol

5 moles of O₂ can be used to produce 3 moles of CO₂.
No. of moles of CO₂ formed = 0.02076 × (3/5) = 0.01246 mol

For CO₂:
PV = nRT
V = nRT/P

Volume of CO₂, V
= nRT/P
= 0.01246 × 0.082 × (273 + 37.3) / 0.973
= 0.326 L
= 326 mL


Method 2 :

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

At STP, 5 volume of O₂ produces 3 volumes of CO₂.
At STP, volume of CO₂ produced = 465 × (3/5) = 279 mL

For CO₂ :
At STP : P₁ = 1.00 atm, V₁ = 279 mL, T₁ =273
Now : P₂ = 0.973 atm, V₂ = ? mL, T₂ = 273 + 37.3 = 310.3 K

P₁V₁/T₁ = P₂V₂/T₂

Volume of CO₂ at the given conditions, V₂
= V₁ × (T₂/T₁) × (P₁/P₂)
= 279 × (310.3/273) × (1.00/0.973) mL
= 326 mL

In what reaction? Without knowing the reaction, we can't help.

Basically, use the stoichiometry of the reaction to calculate the volume of CO2 measured at STP. Then use the equation:
P1V1/T1 = P2V2/T2
to calculate the volume measured under these new conditions.