Question

balance the chemical equation below using the smallest possible whole - number stoichiometric coefficients. ch₃(ch₂)₅ch₃(l) + o₂(g) → co₂(g) + h₂o(g)

Explanation:

Step1: Count atoms on reactant side

The reactant is $CH_3(CH_2)_5CH_3$, which has 7 carbon atoms and 16 hydrogen atoms. Oxygen comes from $O_2$.

Step2: Balance carbon atoms

To balance the 7 carbon atoms in the hydrocarbon on the reactant - side, we put a 7 in front of $CO_2$ on the product - side: $CH_3(CH_2)_5CH_3(l)+O_2(g)
ightarrow7CO_2(g)+H_2O(g)$.

Step3: Balance hydrogen atoms

There are 16 hydrogen atoms in the hydrocarbon. To balance them, we put an 8 in front of $H_2O$ on the product - side: $CH_3(CH_2)_5CH_3(l)+O_2(g)
ightarrow7CO_2(g)+8H_2O(g)$.

Step4: Balance oxygen atoms

On the product - side, there are $7\times2 + 8\times1=22$ oxygen atoms. So we put 11 in front of $O_2$ on the reactant - side.
The balanced equation is $CH_3(CH_2)_5CH_3(l)+11O_2(g)
ightarrow7CO_2(g)+8H_2O(g)$.

Answer:

$CH_3(CH_2)_5CH_3(l)+11O_2(g)
ightarrow7CO_2(g)+8H_2O(g)$