One kilogram of butane (C4H10) is burned with 25 kg of air that is at 308C and 90 kPa. Assuming that the combustion is complete and the pressure of the products is 90 kPa, determine (a) the percentage of theoretical air used and (b) the dew-point temperature of the products.

Explanation:

Combustion of butane gives following reaction

C4H10 + 13/2 O2 → 5H20 + 4CO2

Therefore, One Kg of Butane consists of 1000/58 moles . That is 17.24 moles.

17.24 moles of butane reacts with (6.5×17.24) moles of O2 = 112.06 moles = 3.586 Kg O2.

Weight % of O2 in Air = 20 %

Mass of theoretical air used is 3.586×5 = 17.93 Kg = 71.72 %

The dew point temperature of H2O is :

D = (237.3×B)/(1-B)

B = ln(E/6.108)/17.27

E is vapor pressure at given temp.

At T = 30° C , E = 31.8 mm of Hg = 41.84 Milli Bars

B = 0.1114

D = 29.74°C

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

fe 55.85 g/mol and h2o 18.02 g/mol.