Question

what is the coefficient in front of the o₂·when the equation below is balanced?
c₂h₆(g) + o₂(g) → co₂(g) + h₂o(g)$
use only whole numbers and the lowest whole - number ratio.

Explanation:

Step1: Balance carbon atoms

There are 2 carbon atoms in \(C_2H_6\). So, put a 2 in front of \(CO_2\):
\(C_2H_6(g)+O_2(g)\to2CO_2(g)+H_2O(g)\)

Step2: Balance hydrogen atoms

There are 6 hydrogen atoms in \(C_2H_6\). So, put a 3 in front of \(H_2O\):
\(C_2H_6(g)+O_2(g)\to2CO_2(g)+3H_2O(g)\)

Step3: Balance oxygen atoms

On the right - hand side, there are \(2\times2 + 3\times1=7\) oxygen atoms. So, put \(\frac{7}{2}\) in front of \(O_2\):
\(C_2H_6(g)+\frac{7}{2}O_2(g)\to2CO_2(g)+3H_2O(g)\)
Since we need whole - number coefficients, multiply the entire equation by 2:
\(2C_2H_6(g)+7O_2(g)\to4CO_2(g)+6H_2O(g)\)

Answer:

7