Question #b6258

1 Answer
Oct 25, 2015

Answer:

#C_9H_18# + #27/2O_2# = #9CO_2# + #9H_2O#

or

#2C_9H_18# + #27O_2# = #18CO_2# + #18H_2O#

Explanation:

Separate each element and tally the atoms based on the corresponding subscripts:

Left side:
C = 9
H = 18
O = 2

Right side:
C = 1
H = 2
O = 2 + 1 (DO NOT ADD IT UP YET)

Start with the element that is easiest to balance. In this case, the C.

Left side:
C = 9
H = 18
O = 2

Right side:
C = 1 x 9 = 9
H = 2
O = (2 x 9) + 1

#C_9H_18# + #O_2# = #9CO_2# + #H_2O#

Notice that since #CO_2# is a substance, you also have to multiply the O by 9.

Then go to the next element that is also easy to balance, the H atom.

Left side:
C = 9
H = 18
O = 2

Right side:
C = 1 x 9 = 9
H = 2 x 9 = 18
O = (2 x 9) + (1 x 9)

Again, notice that since #H_2O# is a substance, you need to multiply the O by 9.

#C_9H_18# + #O_2# = #9CO_2# + #9H_2O#

Now the only atom left to balance is the O. Since the total number of oxygen on the right side of an equation is an odd number, you can use your knowledge in fractions to balance

Left side:
C = 9
H = 18
O = #cancel 2# x #27/(cancel 2)# = 27

Right side:
C = 1 x 9 = 9
H = 2 x 9 = 18
O = (2 x 9) + (1 x 9) = 27

Hence, the balanced (reduced) equation is

#C_9H_18# + #27/2O_2# = #9CO_2# + #9H_2O#

Now, if you want whole integers instead of fractions, you can always multiply the whole equation by 2.

#cancel 2# [#C_9H_18# + #27/(cancel 2)O_2# = #9CO_2# + #9H_2O#]

=

#2C_9H_18# + #27O_2# = #18CO_2# + #18H_2O#