An electric discharge is passed through a mixture containing 50 c.c. of $O_{2}$ and 50 c.c. of $H_{2}$ . The volume of the gases formed (i) at room temperature and (ii) at 110⁰C will be

(i) 25 c.c. (ii) 50 c.c.

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(i) 50 c.c. (ii) 75 c.c.

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(i) 25 c.c. (ii) 75 c.c.

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(i) 75 c.c. (ii) 75 c.c.

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Solution

The correct option is C (i) 25 c.c. (ii) 75 c.c.
At room temperature $\begin{matrix} 2 H_{2 (g)} + & O_{2 (g)} \to & 2 H_{2} O_{l} t = 0 & 50 m l & 50 m l & 0 t = t & 50 - 2 x & 50 - x & 2 x = 0 & 25 g a s e s (50) l i q u i d \end{matrix}$
In this case $H_{2}$ is limiting reagent
$x = 25 m l$
At $110^{\circ} C$ $\begin{matrix} 2 H_{2 (g)} + & O_{2 (g)} \to & 2 H_{2} O_{(g)} & V_{g a s} = 75 m l t = t & 0 & 25 m l & 2 \times 50 m l \end{matrix}$

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