a.) what volume of water vapor measured at 250 degrees celsius and 1.00atm is produced when 3.00L of propane at STP is burned?

A.)
The molecular formula of propane is C₃H₈.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

At STP, 1 volume of C₃H₈ produces 4 volumes of water vapor.
At STP, volume of CO₂ produced = 3.00 × 3 = 9.00 L

For water vapor :
At STP : P₁ = 1.00 atm, V₁ = 9.00 L, T₁ =273
Now : P₂ = 1.00 atm, V₂ = ? L, T₂ = 273 + 250 = 523 K

As P₁ = P₂ :
V₁/T₁ = V₂/T₂

Volume of water vapor at the given conditions, V₂
= V₁ × (T₂/T₁)
= 9.00 × (523/273) L
= 17.2 L


B.)
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

At STP, 5 volume of O₂ produces 3 volumes of CO₂.

Volume of O₂ = 642 × (5/3) = 1070 L


C.)
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

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