A 0.430 mol n-butane/mol, 0.570 mol propane/mol mixture at 25°C enters a furnace

Question

A 0.430 mol n-butane/mol, 0.570 mol propane/mol mixture at 25°C enters a furnace. In a separate stream, enters the furnace in 10.0% excess at 341.0°C. The fuel reacts completely. Calculate the adiabatic flame temperature by following the steps below. Take 100 mol of fuel as the basis. Assume the heat capacity for O2 is 3.31 x102 kJ/(mol. C), N2 is 3.13 x 102 kJI(mol. °C), H20 is 3.85 x 102kJ/(mol. "C), and CO2 is 5.00 x 102 kJ/(mol. "C). n1 100 mol fuel yp, 0.570 mol propane/mol yo, 0.430 mol n-butane/mol ng mol exhaust Furnace nsco2 mol CO2 n2 mol air ngH20 mol H20 0.21 mol O2 mol ngo2 mol O2 0.79 mol Namol n3N2 mol N2 What is n2, the moles of air entering the furnace? Number mol air What is nON2, the moles of nitrogen in the exhaust? Number mol N.

Explanation / Answer

We have taken 57 mol of C3H8 and 43 mol of C4H10

The balanced chemical reaction is

C3H8 + 5 O2 --- > 3 CO2 + 4 H2O

Hence stoichiometric amount of O2 required by C3H8 = 57 mol C3H8 x (5 mol O2 / 1 mol C3H8) = 285 mol O2

C4H10 + 6.5 O2 ---- > 4 CO2 + 5 H2O

Hence stoichiometric amount of O2 required by C4H10 = 43 mol C4H10 x (6.5 mol O2 / 1 mol C3H8)

= 279.5 mol O2

Hence total stoichiometric amount of O2 required by the process = 285 + 279.5 = 564.5 mol O2

We have taken 10.0 % excess air.

Hence moles of O2 in air = 564.5 mol O2 x 1.1 = 620.95 mol O2

Moles of N2 in air = (79 / 21) x moles of O2 = (79/21) x 620.95 mol O2 = 2335.95 mol N2

Hence moles of air entering the furance, n2 = moles of O2 + moles of N2

= 620.95 mol + 2335.95 mol = 2957 mol = 2960 mol (upto 3 SF) (answer)

Since N2 doesn't react during the reaction, moles of N2 in the exhaust air, n3N2 =

2335.95 mol N2 = 2340 mol N2 (answer)

Moles of CO2 formed by C3H8 =  57 mol C3H8 x (3 mol O2 / 1 mol C3H8) = 171 mol CO2

Moles of CO2 formed by C4H10 = 43 mol C4H10 x (4 mol O2 / 1 mol C4H10) = 172 mol CO2

Hence total mole of CO2 in the exhaust air, n3CO2 = 171 mol + 172 mol = 343 mol CO2 (answer)

Moles of H2O formed by C3H8 =  57 mol C3H8 x (4 mol H2O / 1 mol C3H8) = 228 mol H2O

Moles of H2O formed by C4H10 = 43 mol C4H10 x (5 mol H2O / 1 mol C4H10) = 215 mol H2O

Hence total mole of CO2 in the exhaust air, n3H2O = 228 mol + 215 mol = 443 mol H2O (answer)

Moles of O2 in the exhaust furance, n3O2 = excess O2 = 564.5 mol O2 x 0.1 = 56.4 mol O2 (answer)

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