Question

current attempt in progress
the balanced equation for the complete combustion of octane, c8h18, would contain:
1c8h18
8co2
25o2
9h2o

Explanation:

Step1: Write the un - balanced equation

The general equation for the combustion of a hydrocarbon \(C_xH_y\) is \(C_xH_y+O_2
ightarrow CO_2 + H_2O\). For octane \(C_8H_{18}\), the un - balanced equation is \(C_8H_{18}+O_2
ightarrow CO_2 + H_2O\).

Step2: Balance the carbon atoms

There are 8 carbon atoms in \(C_8H_{18}\). So, if we have \(CO_2\) as a product, to balance the carbon atoms, we put 8 in front of \(CO_2\). The equation becomes \(C_8H_{18}+O_2
ightarrow8CO_2 + H_2O\).

Step3: Balance the hydrogen atoms

There are 18 hydrogen atoms in \(C_8H_{18}\). In \(H_2O\), since each \(H_2O\) molecule has 2 hydrogen atoms, to balance the hydrogen atoms, we put 9 in front of \(H_2O\). The equation is \(C_8H_{18}+O_2
ightarrow8CO_2 + 9H_2O\).

Step4: Balance the oxygen atoms

On the right - hand side, the number of oxygen atoms in \(8CO_2\) is \(8\times2 = 16\) and in \(9H_2O\) is \(9\times1=9\). So, the total number of oxygen atoms on the right - hand side is \(16 + 9=25\). On the left - hand side, if we let the coefficient of \(O_2\) be \(n\), then \(2n = 25\). But we usually multiply the entire equation by 2 to get rid of the fraction.

Multiplying the equation \(C_8H_{18}+O_2
ightarrow8CO_2 + 9H_2O\) by 2 gives \(2C_8H_{18}+25O_2
ightarrow16CO_2 + 18H_2O\).

If we consider the non - multiplied form (by 2) for the purpose of checking the options:

• For \(C_8H_{18}\), if we take the non - multiplied (by 2) form of the balanced equation \(C_8H_{18}+\frac{25}{2}O_2

ightarrow8CO_2 + 9H_2O\) (we can also work with the non - integer coefficient for \(O_2\) when just checking atom counts per molecule of \(C_8H_{18}\))

• The number of \(H_2O\) molecules: From the hydrogen balance (18 H atoms in \(C_8H_{18}\) giving \(9H_2O\) molecules as \(18\div2 = 9\))

Answer:

\(9H_2O\)