How do we represent the stoichiometric combustion of octane? How much carbon dioxide will result if #15.4*g# of octane are combusted?

1 Answer
Mar 16, 2017

Answer:

We have the stoichiometric equation:

#C_8H_18+25/2O_2 rarr 8CO_2uarr + 9H_2O#

Explanation:

And the given stoichiometry assures us that 8 equiv of carbon dioxide will result for each equiv of octane combusted:

#"Moles of octane"=(15.4*g)/(114.23*g*mol^-1)=0.135*mol#.

And thus #0.135xx8# #"mol"# #CO_2# will be evolved, and this represents a mass of :

#0.135*molxx8xx44.01*g*mol^-1=47.5*g# #"carbon dioxide"#

How much water will be evolved. Will the mass of products equal the mass of reactants? Why?