1. Write the reaction of ethanamine with nitrous acid

Ethanamine is a primary aliphatic amine: $\mathrm{C_2H_5NH_2}$

With $\mathrm{NaNO_2/HCl}$, nitrous acid ($\mathrm{HNO_2}$) is generated in situ, and primary aliphatic amines react as: $\mathrm{RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O}$

For ethanamine: $\mathrm{C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2 + H_2O}$

2. Account for HCl consumed and regenerated

$\mathrm{NaNO_2/HCl}$ means $\mathrm{HCl}$ is used to generate nitrous acid: $\mathrm{NaNO_2 + HCl \rightarrow HNO_2 + NaCl}$

Thus, for each mole of ethanamine reacting, 1 mole of HCl is consumed to form $\mathrm{HNO_2}$.

Now, the question says that after reaction and hydrolysis, $\mathrm{N_2}$ and $\mathrm{HCl}$ are liberated. This corresponds to the standard diazotization representation in acid medium: $\mathrm{C_2H_5NH_2 + NaNO_2 + HCl \rightarrow C_2H_5OH + N_2 + NaCl + H_2O}$

The problem directly states that the HCl generated was completely neutralised by $0.2$ mol of $\mathrm{NaOH}$.

Neutralisation is: $\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$

So, $\text{moles of HCl} = \text{moles of NaOH} = 0.2$

3. Relate moles of HCl to moles of ethanamine

From the reaction stoichiometry, 1 mole of ethanamine corresponds to 1 mole of HCl involved/generated as indicated by the problem statement.

Hence, moles of ethanamine: $n = 0.2 \text{ mol}$

4. Calculate mass of ethanamine

Molar mass of ethanamine, $\mathrm{C_2H_7N}$: $2\times 12 + 7\times 1 + 14 = 24 + 7 + 14 = 45\,\text{g mol}^{-1}$

Therefore, $X = nM = 0.2 \times 45 = 9\,\text{g}$

5. Final answer

$\boxed{9}$