How many grams of barium hydride must be treated with water to obtain 4.36L of hydrogen at `20^(@)C` and 0.975 atm pressure (Ba=137)?

Options

A. 12.28 g
B. 24.56 g
C. 16.14 g
D. 14.56 g

Answer

B

Explanation

`BaH_(2)+2H_(2)O rarr Ba(OH)_(2)+H_(2)`
Pv = nRT
`n=(Pv)/(RT)=(0.975 times 4.36)/(0.0821 times 293)`
1 mole `H_(2)` is obtained from =139 gr `BaH_(2)`
Hence 0.1767 mole `H_(2)` is obtained from
`=139 times 0.1767`
`=24.56gBaH_(2)`