Question

When a metal $\mathrm{X}$ is treated with cold water, it gives a basic salt $\mathrm{Y}$ with molecular formula $\mathrm{XOH}$ (Molecular mass = 40) and liberates a gas $\mathrm{Z}$ which easily catches fire. Identify $\mathrm{X}$, $\mathrm{Y}$ and $\mathrm{Z}$ and also write the reaction involved.

Answer 1

Identifying $\mathrm{X}$, $\mathrm{Y}$ and $\mathrm{Z}$:

1. According to the question as mass of $\mathrm{XOH}$ is 40 and also $\mathrm{X}$ when treated with cold water gives basic salt, so $\mathrm{X}$ is $\mathrm{Na}$ as the molecular mass of $\mathrm{NaOH}$ is 40 and also it is basic in nature.
2. So as $\mathrm{X}$ is $\mathrm{Na}$ and it is giving a basic salt as $\mathrm{Y}$ which would be $\mathrm{NaOH}$.
3. Reaction for this would be: $\underset{\mathrm{Sodium}}{2\mathrm{Na}\left(\mathrm{s}\right)}+\underset{\mathrm{Cold}\mathrm{water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Sodium}\mathrm{hydroxide}}{2\mathrm{NaOH}\left(\mathrm{aq}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$
4. As the gas released here is ${\mathrm{H}}_2$ which will easily catch fire.
5. So $\mathrm{Z}$ would be ${\mathrm{H}}_2$.

Therefore, $$\begin{gathered} \mathrm{X}=\mathrm{Na} \\ \mathrm{Y}=\mathrm{NaOH} \\ \mathrm{Z}={\mathrm{H}}_2 \end{gathered}$$.