How many grams of barium hydride must be treated with water in order to obtain 4.36 L of hydrogen at 20 $^{\circ} C$ and 0.975 atm pressure ?

24.56

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34.56

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12.28

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43. 65

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Solution

The correct option is C $12.28$ g
Moles of $H_{2} = \frac{P V}{R T} = \frac{0.975 \times 4.36}{0.0821 \times 293} = 0.1767$ $m o l e s$
$B a H_{2} + 2 H_{2} O ⟶ B a (O H)_{2} + 2 H_{2}$
$1$ $m o l e$ of $B a H_{2}$ liberates $2$ $m o l e$ $H_{2}$
Moles of $B a H_{2}$ reacted $= \frac{0.1767}{2} = 0.08835$
Mass of $B a H_{2}$ required $= 139 \times 0.08835 = 12.28$ $g m$

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