Question #cf926

2 Answers
Nov 14, 2016

Answer:

Balance the chemical equation!

Explanation:

All you really have to do here is balance the chemical equation that describes the combustion of octane, #"C"_8"H"_18#.

As you know, the complete combustion of octane involves burning this hydrocarbon in the presence of excess oxygen, #"O"_2#. Since octane is a hydrocarbon, i.e. a compound that contains only carbon and hydrogen, the reaction will produce two products

  • carbon dioxide, #"CO"_2#
  • water, #"H"_2"O"#

The unbalanced chemical equation that describes the combustion of octane will thus be

#"C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> "CO"_ (2(g)) + "H"_ 2"O"_ ((l))#

Now just balance this equation like you would fo any chemical equation.

Start with the atoms of carbon. You have #8# present on the reactants' side, so multiply the carbon dioxide by #8# to get #8# atoms of carbon on the products' side

#"C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + "H"_ 2"O"_ ((l))#

Now look at the atoms of hydrogen. You have #18# on the reactants' side and #2# on the products' side, so multiply the water molecule by #9# to get

#"C"_ 8"H"_ (18(l)) + "O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O"_ ((l))#

Finally, focus on the atoms of oxygen. You have #2# on the reactants' side and

#overbrace(8 xx "2 atoms O")^(color(blue)("from 8CO"_2)) + overbrace(9 xx "1 atom O")^(color(purple)("from 9H"_2"O")) = "25 atoms O"#

At this point, you can use a little trick to make the balancing easier. Notice that you can add a Fractional coefficient to #"O"_2# to get

#"C"_ 8"H"_ (18(l)) + 25/2"O"_ (2(g)) -> 8"CO"_ (2(g)) + 9"H"_ 2"O"_ ((l))#

The reactants' side now has

#25/2 xx "2 atoms O" = "25 atoms O"#

the same number of atoms of oxygen as the products' side. To get rid of the fractional coefficient, simply multiply all the coefficients by #color(red)(2)#

#color(red)(2)"C"_ 8"H"_ (18(l)) + (25/2 xx color(red)(2))"O"_ (2(g)) -> (8 xx color(red)(2))"CO"_ (2(g)) + (9 xx color(red)(2))"H"_ 2"O"_ ((l))#

The balanced chemical equation will thus be

#2"C"_ 8"H"_ (18(l)) + 25"O"_ (2(g)) -> 16"CO"_ (2(g)) + 18"H"_ 2"O"_ ((l))#

Now simply add the coefficients to get their sum

#sum"coefficients" = 2 + 25 + 16 + 18 = color(darkgreen)(ul(color(black)(61)))#

Nov 14, 2016

I shall first try to find out a general formula to know the sum of the coefficients of reactants and products for the combustion raction of any alkane.

The general molecular formula of an alkane is #C_xH_(2x+2)# and the blanced equation of its combustion reaction is as follows

#C_xH_(2x+2)+1/2(3x+1)O_2->xCO_2+(x+1)H_2O#

So the sum of the coefficients becomes

#S=1+1/2(3x+1)+x+(x+1)#

#=>S=1+2x+1/2(3x+1+2)#

#=>S=1+2x+3/2(x+1)#

It is obvious from above relation that the Sum (S) will always be a whole number if x is odd

Otherwise to get the sum as whole number for even value of x we are to multiply the value of S by 2

In case of octane x = 8.

So the required sum is

#=2xxS=2(1+2x+3/2(x+1))#

#=2(1+2xx8+3/2(8+1))=2+32+27=61#