Answer: The mass of
liberated is 0.0383 g and the mass of NaH that reacted 0.455 g.
Explanation:
The given data is as follows.
Volume = 0.501 L,
Temperature of gas =
= (35 + 273) K = 308 K,
Total pressure = 760 mm Hg
Vapor pressure of water at
= 422 mm Hg
Partial pressure of
= (760 - 422) mm Hg
= 717.8 mm Hg
The chemical equation is as follows.

Using ideal gas equation, we will find the moles of hydrogen gas as follows.
PV = nRT
n = 
As we know,
1 atm = 760 mm Hg
717.8 mm Hg =
= 0.944 atm
Putting the given values into the above formula we will calculate the number of moles as follows.
n = 
= 
= 0.0190 moles
Now, we will find the mass of
as follows.
Mass = Moles × Molar mass
= 0.0190 moles × 2.01588 g/mol
= 0.0383 g
Therefore, mass of hydrogen is 0.0383 g.
As the ratio between
and NaH is 1:1. So, we will calculate the moles of NaH as follows.
Mass = Moles × Molar mass
= 0.0190 moles × 23.99771 g/mol
= 0.455 g
Therefore, mass of NaH is 0.455 g.