Given volume of CO₂ = 89.6 dm³.

At standard temperature and pressure (STP), 1 mole of any gas occupies $22.4 \text{ dm}^3$.
Given volume of $\text{CO}_2 = 89.6 \text{ dm}^3$.

$$n_{\text{CO}_2} = \frac{\text{Volume of } \text{CO}_2}{\text{Molar volume at STP}}$$ $$n_{\text{CO}_2} = \frac{89.6 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 4 \text{ mol}$$

The molar mass of $\text{H}_2\text{O}$ is $2 \times 1.008 \text{ g/mol} (\text{H}) + 16.00 \text{ g/mol} (\text{O}) = 18.016 \text{ g/mol} \approx 18 \text{ g/mol}$.
Given mass of $\text{H}_2\text{O} = 54 \text{ g}$.

$$n_{\text{H}_2\text{O}} = \frac{\text{Mass of } \text{H}_2\text{O}}{\text{Molar mass of } \text{H}_2\text{O}}$$ $$n_{\text{H}_2\text{O}} = \frac{54 \text{ g}}{18 \text{ g/mol}} = 3 \text{ mol}$$

From the complete combustion of a hydrocarbon, all carbon atoms form $\text{CO}_2$ and all hydrogen atoms form $\text{H}_2\text{O}$.
Each mole of $\text{CO}_2$ contains 1 mole of C atoms.
Moles of C atoms = $n_{\text{CO}_2} = 4 \text{ mol}$

Each mole of $\text{H}_2\text{O}$ contains 2 moles of H atoms.
Moles of H atoms = $2 \times n_{\text{H}_2\text{O}} = 2 \times 3 \text{ mol} = 6 \text{ mol}$

The ratio of carbon atoms to hydrogen atoms in the hydrocarbon is $4:6$.
This simplifies to $2:3$.
So, the empirical formula is $\text{C}_2\text{H}_3$.

Let the general formula of the hydrocarbon be $\text{C}_x\text{H}_y$.
From the moles of C and H atoms, we have $x=4$ and $y=6$.
Therefore, the molecular formula of the hydrocarbon is $\text{C}_4\text{H}_6$.

The general formula for alkanes is $\text{C}_n\text{H}_{2n+2}$. For $n=4$, this would be $\text{C}_4\text{H}_{10}$.
The general formula for alkenes is $\text{C}_n\text{H}_{2n}$. For $n=4$, this would be $\text{C}_4\text{H}_8$.
The general formula for alkynes or dienes is $\text{C}_n\text{H}_{2n-2}$. For $n=4$, this would be $\text{C}_4\text{H}_{2(4)-2} = \text{C}_4\text{H}_6$.

Thus, the hydrocarbon is an alkyne or a diene.