04 question (1 point) write the chemical equation describing the complete combustion of liquid octane…

Explanation:

Step1: Recall combustion reaction principle

In complete - combustion of a hydrocarbon, it reacts with oxygen ($O_2$) to produce carbon dioxide ($CO_2$) and water ($H_2O$).

Step2: Write the un - balanced equation

The reactants are octane ($C_8H_{18}(l)$) and oxygen ($O_2(g)$), and the products are carbon dioxide ($CO_2(g)$) and water ($H_2O(g)$). The un - balanced equation is $C_8H_{18}(l)+O_2(g)\rightarrow CO_2(g)+H_2O(g)$.

Step3: Balance carbon atoms

There are 8 carbon atoms in octane. So, we put a coefficient of 8 in front of $CO_2$: $C_8H_{18}(l)+O_2(g)\rightarrow8CO_2(g)+H_2O(g)$.

Step4: Balance hydrogen atoms

There are 18 hydrogen atoms in octane. So, we put a coefficient of 9 in front of $H_2O$: $C_8H_{18}(l)+O_2(g)\rightarrow8CO_2(g)+9H_2O(g)$.

Step5: Balance oxygen atoms

On the right - hand side, there are $8\times2 + 9\times1=16 + 9 = 25$ oxygen atoms. So, we put a coefficient of $\frac{25}{2}$ in front of $O_2$: $C_8H_{18}(l)+\frac{25}{2}O_2(g)\rightarrow8CO_2(g)+9H_2O(g)$.

Step6: Multiply through by 2 to get whole - number coefficients

$2C_8H_{18}(l)+25O_2(g)\rightarrow16CO_2(g)+18H_2O(g)$

Answer:

$2C_8H_{18}(l)+25O_2(g)\rightarrow16CO_2(g)+18H_2O(g)$