Lithium reacts with water to produce hydrogen gas and lithium hydroxide. What volume of hydrogen collected over water at 22° C and 750 mm Hg pressure is produced by the reaction of 0.208 g of Li? $[{VP}_{H_{2} O} = 19.8 mmHg] [Li = 7]$

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360 mL

755 mL

735 mL

378 mL

answer is B.

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Detailed Solution

$\begin{array}{l} Li + H_{2} O ⟶ LiOH + \frac{1}{2} H_{2} \\ 0.03 mol \frac{0.03}{2} mol \\ \frac{0.03}{2} = \frac{\frac{(750 - 19.8)}{760} \times V}{0.0821 x 295} \\ ∴ V = 378 mL \end{array}$

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