Answer to Question #158906 in Organic Chemistry for Annie

Question

Propane is the major constituent of LPG. 

The enthalpy change for the reaction of propane with oxygen

C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)

is –220 kJ mol^-1.

At 273 K and 101.3 kPa, the volume of carbon dioxide formed per kilojoule of energy released is:

Answer 1

$C_3H_{8 (g) } + 5 O_{2 (g) }\to 3 CO_{2 (g)} + 4 H_2O_{ (g)} \ -220\textsf{kJ/mol}$

from the reaction above,

220kj of heat = 3 moles of CO₂ = 67.2dm³ of CO₂

1 kj of heat = x dm³ of CO₂

x = $\dfrac{67.2}{220} \ L$ = 0.305 L