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Ideal Gas Equation

0.01 mole of ...

Question

$0.01 m o l e$ of iodoform $(C H I_{3})$ reacts with excess $A g$ powder to produce a gas, whose volume at NTP is:

224 m l

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112 m l

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336 m l

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448 m l

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Solution

The correct option is B $112 m l$
The reaction would be $2 C H I_{3} + 6 A g \to 6 A g I + C_{2} H_{2}$

$2 m o l e s C H I_{3}$ produce $1 m o l e$ ethylene gas.

$0.01 m o l e$ will produce $= \frac{0.01}{2} = 0.005 m o l e$ gas

Volume of $0.005$ mole gas at NTP $= 0.005 \times 22.4 = 0.112 L = 112 m l$

Volume of gas $(C_{2} H_{2}) = 112 m l$

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