At electric discharge is passed through a mixture containing 50 cc of $O_{2}$ and 50 cc of $H_{2}$ . The volume of the gas removed at room temperature will be, respectively:

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Solution

$2 H_{2} + O_{2} ⟶ 2 H_{2} O$

$50 c c$ of $H_{2}$ will combine with $25 c c$ of $O_{2}$ to form $50 c c$ $H_{2} O$

$O_{2}$ left = $25 c c$

$∴$ Volume of gas reacted $= (50 + 25) = 75 c c$ .

So, the volume of the gas removed at room temperature will be $25 c c$ of $O_{2}$ and $50 c c$ of $H_{2}$ .

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At electric discharge is passed through a mixture containing 50 cc of O2and 50 cc of H2. The volume of the gas removed at room temperature will be, respectively:

2H2+O2⟶2H2O50cc of H2 will combine with 25cc of O2 to form 50cc H2OO2 left = 25cc∴ Volume of gas reacted =(50+25)=75cc .So, the volume of the gas removed at room temperature will be 25cc of O2 and 50cc of H2.

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