Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 58. g of octane is mixed with 133. g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

2C6H14 + 19O2 ===> 12CO2 + 14H2O ... balanced equation

moled hexane present = 2.6 g x 1 mole/130 g = 0.02 moles

moles O2 present = 5.29 g x 1 mole/32 g = 0.165 moles

Which reactant is limiting? Hexane = 0.02/2 = 0.01; O2 = 0.165/19 = 0.0087

Thus O2 is limiting...

moles of H2O that can be produced =0.165 moles O2 x 14 H2O/19 CO2 = 0.122 moles H2O

Mass H2O = 0.122 moles x 18 gm/mole = 2.20 g (to 3 sig. figs.)