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

h let me think

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.