How do you balance #C_2H_4 + O_2 -> CO_2 + H_2O#?

1 Answer
Dec 25, 2015

Answer:

#color(red)(1)C_2H_4 + color(red)(3)O_2 rarr color(red)(2)CO_2+color(red)(2)H_2O#

Explanation:

Set the proportions so that:
#color(white)("XXX")color(red)(p)C_2H_4+color(red)(q)O_2 rarr color(red)(r)CO_2+color(red)(s)H_2O#

C: [1]#color(white)("XXX")2p - r = 0#
H: [2]#color(white)("XXX")4p-2s=0#
O: [3]#color(white)("XXX")2q-2r-s=0#

We have 3 equations in 4 unknowns which is insufficient to give a unique solution (but then we shouldn't have expected one).

We know that we want #p, q, r, s in ZZ#
but if we temporarily relax this requirement we can determine relative proportions by setting one of them to a a constant and solving for the other values.

I (arbitrarily) chose #p=1#

from [1] with #p=1#
[4]#color(white)("XXX")r=2#
from [2] with #p=1#
[5]#color(white)("XXX")s=2#
from[3] with #r=2# and #s=2#
[6]#color(white)("XXX")q=3#

So
#color(white)("XXX")p:q:r:s = 1:3:2:2#

(Normally at this point we would have some fractional results and need to multiply through be the LCM of the denominators,
but in this case we got lucky and have the integer values we need directly).