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

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Solution

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

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

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