Question

What is the mass of each product when 150.0 g of pentene (C5H10) reacts with excess oxygen?

Answer 1

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