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Gay Lussac's Law, Avagadro's Law

A sample of w...

Question

A sample of water gas has a composition by volume of 50% $H_{2}$ , 45% $C O$ and 5% $C O_{2}$ . Calculate the volume in litres at STP at which the water gas on treatment with excess of steam will produce 5 L of $H_{2}$ . The equation for the reaction is: $C O + H_{2} O \to C O_{2} + H_{2}$

5.26 L

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2.33 L

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5.00 L

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7.20 L

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Solution

The correct option is A $5.26 L$
Given information:
Composition of the water gas is 50% $H_{2}$ , 45% $C O$ and 5% $C O_{2}$
Let the volume of water gas be x L. Then,
volume of $H_{2}$ = $0.5 x$ L
volume of $C O$ = $0.45 x$ L
volume of $C O_{2}$ = $0.05 x$ L
Final volume of $H_{2}$ after reaction with steam = 5 L

$C O$ in water gas reacts with steam to produce $H_{2}$ according to the equation,

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

Applying Gay Lussac's law,

volume of $H_{2}$ formed = volume of $C O$ consumed = $0.45 x$ L

Final volume of $H_{2}$ =
(volume of $H_{2}$ present initially + volume $H_{2}$ formed from $C O$ )
$\Rightarrow 5 L = 0.5 x L + 0.45 x L$
$\Rightarrow x = 5.26 L$

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Gay Lussac's Law, Avagadro's Law

Standard XII Chemistry

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A sample of water gas has a composition by volume of 50% H2, 45% CO and 5% CO2. Calculate the volume in litres at STP at which the water gas on treatment with excess of steam will produce 5 L of H2. The equation for the reaction is: CO+H2O→CO2+H2

A sample of water gas has a composition by volume of 50% $H_{2}$ , 45% $C O$ and 5% $C O_{2}$ . Calculate the volume in litres at STP at which the water gas on treatment with excess of steam will produce 5 L of $H_{2}$ . The equation for the reaction is: $C O + H_{2} O \to C O_{2} + H_{2}$