✔ 最佳答案
C3H8(g)+ 5O2(g) → 3CO2(g) + H2O(l)
Mole ratio C3H8 : O2 = 1 : 5
For complete ignition of 5 cm³ of C3H8, 250 cm³ of O2is needed.
Hence, O2 is in excess, and C3H8 is thelimiting reactant (completely reacted).
Volume C3H8 left = 0
Volume of O2 left = 300 - 5*50 = 50 cm³
Volume of CO2 formed = 3*50 = 150 cm³
Volume of H2O liquid formed » 0cm³
Volume of resulting gas mixture = 50 + 150 = 200 cm³