Propane is burned completely with excess oxygen. The product gas contains 24.5 m
ID: 495096 • Letter: P
Question
Propane is burned completely with excess oxygen. The product gas contains 24.5 mole% CO_2, 6.10% CO, 40.8% H_2 O, and 28.6% O_2. (a) Calculate the percentage excess O_2 fed to the furnace. (b) A student wrote the stoichiometric equation of the combustion of propane to form CO_2 and CO as 2C_3 H_8 + 17/2 O_2 rightarrow 3CO_2 + 3CO + 8H_2 O According to this equation, CO_2 and CO should be in a ratio of 1/1 in the reaction products, but in the product gas of Part (a) they are in a ratio of 24.8/6.12. Is that result possible?Explanation / Answer
Combustion of propane can be represented by
C3H8+ 5O2----à3CO2 + 4H2O (1)
Also C3H8+ 3.5O2-----à3CO+ 4H2O
Basis : 100 moles of product gas. It contains 24.5 moles CO2, 6.1 moles CO, 40.8 moles H2O, 28.6% O2.
From the reaction, 1 mole of C3H8 produces 3 moles of CO2.
24.5 moles of CO2 required 24.5/3= 8.2 moles of C3H8 . It also produces 6.1 moles of CO which comes from incomplete combustion of CO. This also 6.1/3= 2 moles of C3H8
Total moles of C3H8= 8.2+2= 10.2
Moles of oxygen required as per the reaction-1 =10.2*5 ( 1 mole of C3H8 requires 5 moles of oxygen for formation of CO2). Hence moles of oxygen required= 51 moles. This is the oxygen required for complete combustion.
Oxygen remaining = 28.6 moles
However, 6.1 moles of CO is formed which consumes lesser oxygen than CO2 formed. Hence moles of Oxygen that would have consumed If there is no CO = (6.1/3)*5= 10 moles
Oxygen remained had there been complete conversion of CO2= 28.6-10= 18.6 moles
This has to be excess oxygen.
% excess oxygen = 100*18.6/50=37.2
It is possible, if 50% of C3H8gets converted to CO2 and remaining 50% gets converted to CO, 1; 1 molar ratio is possible. If that is not the case, then the ratio of CO to CO2 can be different.
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