Answer:
66% excess of O₂
Explanation:
a. The combustion of propane (C₃H₈) produce:
C₃H₈ + 5O₂ → 3CO₂ +4H₂O (1)
And:
C₃H₈ + ⁷/₂O₂ → 3CO + 4H₂O (2)
The sum of the two reactions is:
2C₃H₈ + ¹⁷/₂O₂ → 3CO₂ + 3CO + 4H₂O (3)
As 8 moles of H₂O are produced from 2moles of C₃H₈ the initial moles of propane (assuming 100 total initial moles) is:
40,8molH₂O×[tex]\frac{2molC_{3}H_{8}}{8molH_{2}O}[/tex] = 10,2 molesC₃H₈
For a complete reaction of 10,2 moles of C₃H₈ there are necessaries:
10,2 molesC₃H₈×[tex]\frac{17/2molO_{2}}{2molC_{3}H_{8}}[/tex] = 43,35 mol O₂
As the product gas contains 28,6 moles of O₂, the initial total moles are:
43,35mol + 28,6mol = 71,95 initial moles of O₂
The formula of percentage excess O₂ is:
[tex]\frac{InitialMoles-TheoreticalMoles}{TheoreticalMoles}[/tex]×100
Replacing:
[tex]\frac{71,95-43,35}{43,35}[/tex]×100 = 66% excess of O₂
b. The production of CO₂ and CO from C₃H₈ could be seen as two different reactions (1) and (2). The different conditions of combustion of propane could produce more CO₂ over CO and vice versa. For this reason, the ratio of CO₂/CO could be different of 1.
I hope it helps!
