Iso-octane (C_8H_18) is combusted with air which is composed of 21% oxygen and 7

Question

Iso-octane (C_8H_18) is combusted with air which is composed of 21% oxygen and 79% nitrogen (e.g. for every mole of O_2 there arc 3.76 moles of N_2). Assume that combustion occurs under stoichiometric conditions (e.g. the only products of this reaction arc CO_2, H_2O, and N_2). (i) Write an atom balanced chemical reaction equation. (ii) What is the air/fuel ratio on a MOLAR basis? (iii) What is the air/fuel ratio on a MASS basis? (iv) How do answers to (i), (ii), and (iii) change if iso-octane is burned in pure oxygen?

Explanation / Answer

C8H18 + 27/2 O2 ------> 8CO2 + 9H20
1 mole of fuel needs 27/2 moles of O2
1 mole of air contains 21% of O2
21/100 moles of O2 requires 1 mole of air
27/2 moles of O2 requires 21/100*27/2 moles of air
                         = 2.835 moles of air
ratio of air/fuel on a molar basis = 1/2.835
ratio of air/fuel on a mass basis = 2.835*((32*21/100) + (28*79/100))/(1*(8*12+18*1))
                                  = 0.7172
C8H18 burns in pure oxygen then,
ratio of air/fuel on a molar basis = 27/2/1 = 27:2
ratio of air/fuel on a mass basis = 27*32/2*114 = 3.789

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