Hydrogen - Result Question 2
####2. When a substance $A$ reacts with water, it produces a combustible gas $B$ and a solution of substance $C$ in water. When another substance $D$ reacts with this solution of $C$, it also produces the same gas $B$ on warming but $D$ can produce gas $B$ on reaction with dilute sulphuric acid at room temperature. $A$ imparts a deep golden yellow colour to a smokeless flame of Bunsen burner. $A, B, C$ and $D$ respectively are
[1998]
(a) $Na, H_2, NaOH, Zn$
(b) $K, H_2, KOH, Al$
(c) $Ca, H_2, Ca(OH)_2, Sn$
(d) $CaC_2, C_2 H_2, Ca(OH)_2, Fe$
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Solution:
(a) $\underset{\text{ ‘A’ }}{2 Na}+\underset{H_2 O}{2 H_2 O} \to \underset{C^{\prime}}{2 NaOH}+\underset{H_2}{H_2 \uparrow}$
$Zn+2 NaOH \to Na_2 ZnO_2+H_2 \uparrow$
‘D’ ‘C’ ‘B’
$\underset{\text{ ‘D’ }}{Zn}+$ dil. $H_2 SO_4 \to \underset{{ }^{\prime} B^{\prime}}{ZnSO_4}+\underset{H_2}{ }$
$Na$ produces golden yellow colour with smokeless flame of Bunsen burner.
$ \begin{aligned} & \text{ (c) } Fe+\text{ dil. } H_2 SO_4 \to FeSO_4+H_2 \uparrow \\ & 3 Fe+4 H_2 O \to Fe_3 O_4+4 H_2 \uparrow \\ & Steam \\ & Cu+\text{ dil. } HCl \to \text{ No reaction } \\ & \text{ Copper does not evolve } H_2 \text{ from acid as it is } \\ & \text{ below hydrogen in electrochemical series. } \\ & 2 Na+C_2 H_5 OH \to 2 C_2 H_5 ONa^{-} H_2 \uparrow \\ & \text{ (b) } H^{-}(\text{aq })+H_2 O(l) \longrightarrow OH^{-}(aq)+H_2(g) \\ & \text{ base } 1 \text{ acid } 1 \text{ base 2 acid2 } \end{aligned} $
In this reaction $H^{-}$acts as bronsted base as it accepts one proton $(H^{+})$from $H_2 O$ to form $H_2$.
