Can you help me with this sodium hydride reaction problem?

Question

Sodium hydride reacts with excess water to produce aqueous sodium hydroxide and hydrogen gas

 

NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)

 

How do I calculate what mass will produce 982mL of gas at 28°C and 765 torr, when the hydrogen is collected over water. The vapor pressure of the water this temperature is 28 torr?

Answer 1

NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)

You want to calculate the mass of NaH, I assume.  Otherwise, the question isn't clear.  It simply says calculate the mass(??)

 

So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr.  But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas.  So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.

PV = nRT

n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)

n = 0.0385 moles H2

 

moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required

mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH