What coefficient would the #O_2# have after balancing #C_2H_8 + O_2 -> CO_2 + H_2O#?

1 Answer
Feb 9, 2018

Answer:

4

Explanation:

#C_2H_8+O_2rarrCO_2+H_2O#

First, I advise you to make a brief list of values for each side:

LHS

C = 2
H = 8
O = 2

RHS

C = 1
H = 2
O = 3

When balancing chemical equations, we should always try to balance the term with the most elements involved or one with an odd number of atoms. A single element is usually left until the end because it can be balanced without affecting other elements.

In this case, the oxygen on the RHS has only 3. If we multiply by 4, we are able to balance the hydrogen and get the oxygen even.

#C_2H_8+O_2rarrCO_2+4H_2O#

LHS

C = 2
H = 8
O = 2

RHS

C = 1
H = 8
O = 6

Let's multiply the #CO_2# 2 to balance to carbon:

#C_2H_8+O_2rarr2CO_2+4H_2O#

LHS

C = 2
H = 8
O = 2

RHS

C = 2
H = 8
O = 8

Finally, if we balance the oxygen by multiplying the #O_2# by 4, we are done:

#C_2H_8+4O_2rarr2CO_2+4H_2O#

Answer = 4