An electric discharge is passes through a mixture containing 50 c.c. of $H_{2}$ and 50 c.c. of $O_{2}$ . The volume of the gases formed (i) at room temperature and (ii) at $110^{\circ} C$ will be:

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Solution

$2 H_{2} + O_{2} \to 2 H_{2} O$

$2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(l)}$ at room temp

$2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(g)}$ at $110^{\circ} C$

At room temperature 25mL of $O_{2}$ gas would be left unreacted while at $110^{\circ} C$ 50mL of $H_{2} O$ and 25mL of $O_{2}$ will be present.

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An electric discharge is passes through a mixture containing 50 c.c. of H2 and 50 c.c. of O2. The volume of the gases formed (i) at room temperature and (ii) at 110∘C will be:

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An electric discharge is passes through a mixture containing 50 c.c. of $H_{2}$ and 50 c.c. of $O_{2}$ . The volume of the gases formed (i) at room temperature and (ii) at $110^{\circ} C$ will be: