When a metal X is treated with cold water, it | vedclass

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Question

(A) $1$. Given that the molecular formula of the salt $Y$ is $XOH$ and its molecular mass is $40$.$2$. The atomic mass of $O$ is $16$ and $H$ is $1$.

Vedclass · Accepted Answer

$X$ = Na, $Y$ = NaOH, $Z$ = $H_2$

Vedclass · Answer

(A) $1$. Given that the molecular formula of the salt $Y$ is $XOH$ and its molecular mass is $40$.$2$. The atomic mass of $O$ is $16$ and $H$ is $1$. Therefore, $X + 16 + 1 = 40$, which gives $X = 23$. The element with atomic mass $23$ is Sodium $(Na)$.$3$. Thus, $X$ is $Na$, $Y$ is $NaOH$, and the gas $Z$ liberated is Hydrogen $(H_2)$, which is highly flammable.$4$. The balanced chemical equation is: $2Na(s) + 2H_2O(l) ightarrow 2NaOH(aq) + H_2(g) + 	ext{Heat energy}$.

$X$	$Y$
$1$. Extraction of sulphur	$p$. Contact process
$2$. Production of nitric acid	$q$. Frasch method
$3$. Production of sulphuric acid	$r$. Haber's process
$4$. Production of ammonia gas	$s$. Ostwald's method

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 write the chemical reaction involved.

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