Question

Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 72. g of hexane is mixed with 150. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

hexane is C6H14 or put another way, as in the question CH3(CH2)4CH3. Actually, it is incorrect as depicted in the question.

Nonetheless...

2C6H14 + 18O2 ===> 12CO2 + 14H2O ... balanced equation

moled hexane present = 72 g x 1 mole/130 g = 0.553 moles

moles O2 present = 150 g x 1 mole/32 g = 4.68 moles

Which reactant is limiting? Hexane = 0.553/2 = 0.2765; O2 = 4.68/18 = 0.26

Thus O2 is limiting...

moles of H2O that can be produced = 4.68 moles O2 x 14 H2O/18 CO2 = 3.64 moles H2O

Mass H2O = 3.64 moles x 18 gm/mole = 65.52 g (to 3 sig. figs.)