In: Chemistry
For combustion of Pentane first we need a balenced equation as:
C5H12 + 8O2 -----> 5CO2 + 6H2O
so we can say: 72g (MW pentane) petane react with 128g (MW O2) of O2 to produce
220g (MW CO2) of CO2 and 108g (MW H2O) of H2O.
so, If we take 150g of pentane, the products CO2 and H2O will form in the similar proportion as it react with excess of oxyge (means complete concumption of pentane)
wt of pentane for question w = 150g
so, Mass of CO2 = mw CO2 x w / mw pentane
CO2 = 220 x150/72 = 458g
Mass of H2O = mw H2O x w /mw pentane
H2O = 108 x 150 /72 = 225g