`23 g` sodium metal reacts with water. Calculate the:
(a) volume of `H_(2)` liberated at `NTP`
(b) moles of `H_(2)` liberated,
(c ) weight of `H_(2)` liberated.

Explanation

The given reaction is
`2Na+2H_(2)Orarr2NaOH+H_(2)`
From equation, it is evident:
`46 g Na` reacts to liberate `1 "mole" H_(2)`,
`:. 23 g Na` reacts to liberate `(1xx23)//46= 1//2 "mole"H_(2)`
`:.` Weight of `H_(2)` liberated `=(1//2)xx2= 1g`
Also, volume of `H_(2)` at `STP= 22400xx(1)/(2)= 11200 mL`
Alternative method
`{:(,2Na+,2H_(2)Orarr,2NaOH,+H_(2)),("Mole before reaction",23//23=1,"Excess",0,0),("Mole after reaction",0,"Excess",1,1//2):}`
`:.` Moles of `H_(2)` formed `= 1//2`
Wt. of `H_(2)` formed `= (1//2)xx2= 1g`
Volume of `H_(2)` formed at `STP= 22400xx1//2`
`= 11200 mL`